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Table Of Contents
  • Level Of Significance
  • Levene’s Test of Equality of Variance
  • Paired Samples Test

Level Of Significance

The average family size in the US was reported to be 3.15. A random sample of families taken in a local school district is listed in the data file FamilySize.sav located on CANVAS. A local school board member claims that SW Florida has a different family size compared to the rest of the United States. At a 5% level of significance does the local average family size differ from the national average
a. State the null and alternate hypotheses in symbolic form.
Level Of Significance
b. Compute and report the sample size and test statistic.
Ans: Sample size = 53 and test statistic = 1.804.
c. Compute and report the Sig. Value (P-Value):
Ans: P-value = 0.077
d. Make a decision in reference to the null hypothesis:
Ans: Fail to reject the null hypothesis.
e. Summarize the results of the output in context to the claim by the school official:
Ans: There is not sufficient evidence to support the claim that the local average family size differ from the national average.
Output:
One-Sample Statistics
  N Mean Std. Deviation Std. Error Mean
The average size of a household in the United States 53 3.5283 1.52673 .20971
 
One-Sample Test
  Test Value = 3.15
t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference
Lower Upper
The average size of a household in the United States 1.804 52 .077 .37830 -.0425 .7991

Levene’s Test of Equality of Variance

Use the data file titled TuitionCosts located on CANVAS to answer the following question: Is there a difference in tuition costs between public and private universities? A sampling of tuition costs from a random sample of a New England state four-year colleges was obtained for this comparison.
a. State the null and alternate hypotheses in symbolic form:
Level Of Significance 1
b. Using the Levene’s Test of Equality of Variances and the SPSS generated Sig. value; can we say the two groups have equality of variances? (Provide the Sig. Value and answer the question determining if the two groups can be said to have equality of variances).
Ans: The significance value of 0.039, thus we can reject the null hypothesis that the variances are equal. Hence we can say that the variances are not equal.
c. Compute and report the test statistic in the following format t(d.f.):
Ans: 3.224(10.304)
d. Compute and report the Sig. Value (P-Value) for the independent samples t-test:
Ans: The significance value = 0.009.
e. Make a decision in reference to the null hypothesis (reject or fail to reject):
Ans: Reject the null hypothesis.
f. Summarize the results of the output obtained in context to the claim by the official:
Ans: Yes, there is sufficient evidence to support the claim that there is a difference in tuition costs between public and private universities
 Output:
 
Independent Samples Test
  Levene's Test for Equality of Variances                                             t-test for Equality of Means
F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference
Lower Upper
Cost of University tuitions Equal variances assumed 4.930 .039 3.472 18 .003 11366.62626 3273.33361 4489.60753 18243.64500
Equal variances not assumed     3.224 10.304 .009 11366.62626 3525.91594 3541.66332 19191.58921

An educational researcher developed a wooden toy assembly project to test learning in 6-year old children. The time in seconds for the child to assemble the toy was recorded in seconds for a group of 10 children. After assembling the toy, the toy was dissembled out of the sight of the test subjects, then the subjects were again asked to assemble the wooden toy. The researcher stated that if the time to assemble the toy was significantly less the second time, learning had occurred. Using the file, ToyTime.sav to determine the veracity of the researchers claim.
a. State the null and alternate hypotheses in symbolic form.
Level Of Significance 2
b. What is the mean score and standard deviation for the first trial?
Ans: Mean score = 125.9 and standard deviation = 17.078
c. What is the mean score and standard deviation for the second trail?
Ans: Mean score = 110 and standard deviation = 19.528
d. Compute and report the test statistic.
Ans: Test statistic: 4.671(9)
e. Compute and report the SPSS generated Sig. Value to three decimal places,
Ans: Sig. value for two tailed test is 0.001. Thus for left-tailed test it is 0.0005.
f. Make a decision regarding the null hypothesis.
Ans: Reject the null hypothesis.
g. Summarize the results.
Ans: There is sufficient evidence to support the claim that the average score in the second trial is lesser.

 

Paired Samples Test

  Paired Differences t df Sig. (2-tailed)
Mean Std. Deviation Std. Error Mean 95% Confidence Interval of the Difference
Lower Upper
Pair 1 Trial_1 - Trial_2 15.90000 10.76465 3.40408 8.19943 23.60057 4.671 9 .001

Cohen’s Conventions

Compute and report the effect size (Cohen’s d) for question 3. Based upon the value calculated, is this a small, medium, or large effect based upon Cohen’s conventions. (Note: SPSS does not generate this calculation)

Level Of Significance 3