# Inferential Statistics Assignment Solutions

We offer free and accurate inferential statistics assignment solutions below. The solutions touch on sample sizes; confidence intervals; hypothesis testing and errors; and other concepts of inferential statistics.

## Finding Confidence Intervals, Testing Hypotheses, and Verifying Sampling Statements

The following solutions are developed around questions asking about confidence intervals, samples, populations, and other aspects of inferential statistics. While some of them give true/false answers to given statements, others are based on testing hypotheses and finding the confidence intervals.

Answering True/False to Inferential Statistics Questions

Question 1

In practice, it is usually impossible to know the value of a population parameter with complete certainty, but it is usually easy to determine the value of a sample statistic with complete certainty

The reason for the answer is that, if the population is too large, it will be difficult to calculate its population parameter. For example, population means and population variance with a complete reliable result. In contrast, when the population is small, it is easy to calculate with complete certainty.

Question 2

An estimator is a strategy or rule that is used to estimate a population parameter.

An estimator is a statistic used to estimate a population parameter. Due to the large size study of where the population is difficult, to reduce this difficulty, we study the sample i.e. small part of the population and with the help of these statistics, we estimate the population parameter.

Question 3

A point estimator is a single number used to estimate a population parameter.

A point estimator is a single quantity based on the sample used to estimate a parameter. With the help of a single quantity of sample mean we estimate its population mean.

Question 4

An interval estimate contains an upper and lower boundary that will hopefully contain the population parameter.

Confidence interval is the interval formed by upper and lower boundary based on sample data within which parameter will lie with very high probability

Question 5

The sample mean of a simple random sample is a poor estimator of the population mean, but the sample range of a simple random sample is a good estimator of the population range

The sample mean of a simple random sample is a strong unbiased estimator of the population mean. Sample range is not unbiased estimator of population of population range. The range of sample will only be this large if population minimum and maximum value in the distribution are in samples.

Question 6

For a given population, one can achieve greater confidence (in an estimation of the population mean) by either:

1. Increasing the sample size, or
2. Accepting a large margin of error

The general form of the confidence interval is given as

Estimate ± margin error

If the margin of error increased, then the confidence interval becomes wider. That implies more confidence.

Now in the margin of error, in the denominator, there is the square root of sample size. If the sample size is increased, then the denominator will be increased and finally, the margin of error increased. Therefore confidence intervals become wider. That implies more confidence.

Question 7

For a given data sample, an 80% confidence interval will be wider than a 90% confidence interval

From a given data sample, 90 % CI interval contains 80 % CI

Question 8

The confidence interval for a proportion must always be between 0 and 1, inclusive.

It can be negative if the sample proportion is less than the assumed population proportion to be tested

Question 9

In real-world situations of trying to estimate a population mean, it is more realistic and common to Not know a population standard deviation than it is to know the population standard deviation

Realistic if we know population standard deviation

Question 10

In the process of using a simple random sample to approximate the mean of a normally distributed population, it’s advisable to use the z-distribution if the standard deviation is not known, and recommended that you use the t-distribution if the std dev. is specified.

The other way around, for the application of z distribution, population standard deviation should be known for a large sample

For the application of t distribution, if the sample size is small, (< 30), and sample standard deviation should be unknown, and also parent population should be normal

Question 11

The t-distribution only has one parameter, degrees of freedom

For a sample of size n, the t-distribution has (n - 1) degrees of freedom. This is the only parameter required to define t-distribution, and it has no other parameter.

Question 12

As the sample size becomes larger, the shape of the t-distribution approaches the shape of the normal distribution.

The larger the sample size, the more symmetric the t-distribution becomes, asymptotically approaches the normal distribution

Question 13

If a decimal answer result from calculating the sample size needed to achieve a desired margin of error, one should always round UP to the next whole number

While calculating the sample size required to achieve a certain margin of error at a given confidence level, we want to ensure that the margin of error remains within the desired level, never exceeding it. If the sample size calculated under the given condition yields a decimal or fractional value, it should always be rounded up to the next integer and not rounded down even if the general rules of rounding indicate that it should be rounded down. This is because, if it is rounded down, the margin of error will exceed the desired level.

Question 14

In hypothesis testing, the null hypothesis always contains an equality statement of equality, that is, it will contain either >, <, or =

It is correct that the null hypothesis defined in a hypothesis testing problem will always contain a statement of equality so that it must contain one of the symbols, ≥, ≤, or =, and not > or <.

Question 15

In hypothesis testing, the null hypothesis is presumed to be false unless sample data produces overwhelming evidence to the contrary.

Question 16

In hypothesis testing, the results will lead to one of two choices: either “reject the null hypothesis: or “fail to reject the null hypothesis”

The results are mainly of two choices; reject or fail to reject the null hypothesis

Question 17

In hypothesis testing, if you wrongly reject an H0, you have committed a type 1 error

While making a decision, two types of error can be made, either reject the H_0when it is true, or accept the H_0when it is false. When the H_0s are rejected when it is true, such errors are called type 1 errors.

Question 18

While testing a hypothesis, “failing to reject the H0” if the H1 is the correct choice is committing a type 2 error

While making a decision, two types of error can be made, either reject the H_0 when it in fact is true, or accept the H_0when it is false.

When the H_0 can not be rejected when in fact the alternative hypothesis is correct, such errors are called type II error

Question 19

The probability of a Type 1 error can be controlled in the hypothesis testing procedure

In hypothesis testing, from both types of errors, Type 1 error is more serious than Type II error and thus Type 1 error is fixed and then try to reduce Type II error in respect of fixed Type 1 error, thus the Type 1 error can be controlled.

Question 20

The probabilities of committing type I and II errors are inversely proportional. Hence, when the probability of a type I error gets smaller, there are more chances of committing a type II error.

The probabilities of making both types of errors have an inverse proportional relationship.

Finding the Sample Mean

Question 21

Jimbo is trying to estimate a population mean, and he calculates an interval of <24.5, 32.5>, with a 95% confidence level. What is the level of the sample mean?

The Confidence interval for the population mean is given as:

C.I=(x ̅-s/√n t_(α/2,n-1),x ̅+s/√n t_(α/2,n-1) )

Note:

t_(α/2,n-1) is the critical value observed from the t-distribution

Recall that the 95% confidence interval will be (24.5, 32.5)

For the lower limit,

x ̅-s/√n t_(α/2,n-1)=24.5

For the upper limit,

x ̅+s/√n t_(α/2,n-1)=32.5

Solving simultaneously,

2x ̅=57.0

x ̅=28.5

Therefore, the sample mean is x ̅=28.5

Question 22

Jimbo is trying to estimate a population mean, and he calculates an interval of <24.5, 32.5>, with a 95% confidence interval. Which of these is the correct interpretation of these results?

Jimbo is 95% certain that the true population mean lies is equal to or between 24.5 and 32.5.

Constructing the Confidence Interval

Question 23

Construct a 90% CI for the actual mean of a normally distributed population if a random sample size of 40 from the population has a sample mean of 75 and the population has an std dev. of 5

Given that

Sample mean x ̅=75, standard deviation = 5, Confidence interval = 90%, n = 40

Alpha (α)=1-90%=0.1

Z_(α/2)=1.645

90% confidence interval for x ̅

=x ̅±Z_(α/2) × σ/√n

=75±1.645×5/√40

=75±1.3005

=73.6995,76.3005

The 90% confidence interval estimate of the population mean is 73.6995 and 76.3005

Question 24

The Anderson produce Company grows tomatoes in Florida for distribution to various parts of the country. 40 boxes are selected using random sampling for weighing. The mean weight for the selected sample is 33.5 pounds/box, and the standard deviation is 2.1 pounds. Find a 90% CI for the true mean weight of the boxes of tomatoes.

Given that:

n=40,sample mean=33.5,standard deviation=2.1,α=90%=0.1

90% confidence interval for x ̅

=x ̅±t_(α/2,n-1) × σ/√n

Critical values t_(α/2,n-1)=1.685

=33.5±1.685×2.1/√40

=33.5±0.5595

=32.941,34.059

The 90% confidence interval is 32.941,34.059

Question 25

An independent group of foodservice personnel conducted a survey on toping practices in a large metropolitan area. They collected information on the percentage of the bill left as a tip for 25 randomly selected bills. The average tip was 12.3% of the bill with a standard deviation of 2.7%. Construct an interval to estimate the true average tip (as a percent of the bill) with 99% confidence.

Sample mean = 12.3, standard deviation 2.7, sample size = 25, alpha = 99% = 0.001

Critical t_(α/2,df)=t_(0.005,24)=±2.80

99% confidence interval for x ̅

=x ̅±t_(α/2,n-1) × σ/√n

=12.3±2.80×2.7/√25

=33.5±1.512

=10.788,13.812

The 99% confidence interval is 10.788,13.812

Question 26

A technician seeks to estimate the preserving ability of a new brand of additives. Based on a previous test, the time to expiry for this additive was determined to have a standard deviation of 6 days. In order to be 90% confident about the true average time to spoilage, find the sample size that will be needed to approximate the meantime to spoilage with an accuracy of 1 day

Given that,

Z /2 = 1.645

Sample size = n = [Z /2* / E] 2

n = [1.645 * 6 / 1]2

n = 97.4

Sample size = n = 98

Question 27

It was reported that 31% of all people who participated in a survey of recent poll claimed that they loved vegetables. 14% of the participants responded that they loved liver, and 10% of those surveyed did not submit a response because they claimed that they liked everything. The poll was based upon a sample of 1001 people. Assuming that the surveyors chose a random sample, construct a 90% CI for the % of all those who claimed that they loved vegetables.

Given that

Proportion p ̂=0.31, Confidence interval = 90%, n = 1001

Alpha (α)=1-90%=0.1

Z_(α/2)=1.645

90% confidence interval for p ̂

=p ̂±Z_(α/2) √((p(1-p))/n)

=0.31±1.645×√((0.31(1-0.31))/1001)

=0.31±1.645×0.0146

=0.31±0.024

=0.286,0.334

Lower limit is 0.286 and the Upper limit is 0.334

Determining an Appropriate Sample Size for a Study

Question 28

A researcher in Africa believes that 40% of families in the area cannot get access to sufficient drinking water either because the available one is contaminated or because there is no water at all. What is the sample size necessary to approximate the % of those without enough water to within 5%, with 99% confidence?

Given that

Margin error = 5% = 0.05

Level of significance = 1% = 0.01

P = 0.40

Sample size is

n=(Z_(α/2)/E)^2 p (1-p)

=(Z_0.05/0.05)^2 p (1-p)

=(2.5758/0.05)^2 0.40(1-0.4)

=636.9356

=637

The sample size is 637

Testing Hypotheses

Question 29

A random sample of 1000 observations produces a sample mean of 53.5 with a standard deviation of 5.3. Test the hypothesis that the mean is not equal to 55 at the 0.05 significance level.

Given that

x ̅=53.5,standard deviation=5.3,sample size=1000

Null hypothesis: H_0:μ=55

Alternative hypothesis: H_1:μ≠55

Test statistics =((x ) ̅-μ)/(s/√n)

=(53.5-55)/(5.3/√1000)

=(-1.5)/(5.3/31.62277)=-8.95

Critical value Z_(α/2)=±1.96

Decision: Reject the null hypothesis when the test statistic is less than the critical value

Conclusion: Since the test statistic < critical value, we reject the null hypothesis. This implies that there is sufficient evidence that the mean is not equal to 55.

Question 30

A regulatory authority has received complaints that a certain sugar company is underfilling its 5-pound bags of sugar. The authority randomly chooses 750 bags of sugar and measures them to determine their actual weight. The sample mean weight turns out to be 4.80 pounds. The standard deviation is determined to be 0.15 pounds. Is there evidence at 0.01 level that these bags of sugar are indeed underfilled?

Given that

Population mean = 5

Sample mean = 4.80

Standard deviation = 0.15

Alpha = 0.01

Test statistics =((x ) ̅-μ)/(s/√n)

=(4.80-5)/(0.15/√750)

=(-0.20)/0.00547=-36.563

Critical value = 2.33

Decision: Reject the null hypothesis when the test statistic is less than the critical value

Conclusion: Since the test statistic < critical value, we reject the null hypothesis. This implies that the bags are underfilled

Question 31

Ordinarily, when a company recruits a technical staff member, about 25% of the applicants are qualified. However, based on the information in 120 recently received resumes, 18 appear to be technically qualified. Is there overwhelming evidence that the percentage of qualified applicants is less than 25%? Test at the 0.05

Given that:

P = 0.25, n = 120, x = 18, alpha = 0.05

Hypothesis

Null hypothesis: H_0:p=0.25

Alternative hypothesis: H_1:p<0.25

Test statistic

=(p ̂-p)/√((p(1-p))/n)

(0.15-0.25)/√((0.25(1-0.25))/120)

=-0.10/0.04

=-2.53

The p-value for z-distribution at α=0.05 is 0.0057

Decision; when the p-value is then alpha (0.05) reject the null hypothesis

Conclusion: Since the p-value is less than alpha, we reject the null hypothesis and conclude that the percentage of qualified applicants is less than 25%.