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Confidence Interval and t-score Hypothesis using Data Analysis assignment Solution

Confidence Intervals and Hypothesis Testing

1. Construct the following two confidence intervals for µ using only the given information. Answers should be in form x − t √s < µ < x + t √s or something Equivalent, where the t−score corresponds to a two-tail probability of α :

(a) n = 16, x = 23, s = 12, α = 0.05 :

(b) n = 25, x = 23, s = 12, α = 0.25 :

2. Suppose a tutoring company advertises that its clientele average 80% in their math courses. Feeling skeptical, you question a sample of 9 students who used the tutoring service, finding that their average was only 68%, with a standard deviation of 11%:

a) State the null and alternative hypotheses, H0 and H1:

b) Calculate the appropriate t−statistic at the 5% significance level:

c) Estimate the p−value, and describe what it means in the context of this problem:

d) State your conclusions (ie: is the tutoring company necessarily not being truthful? Why would we observe an average of 68% when the stated average was so much greater? Share your thoughts):



a) Here n= 16, thus df= 15. Hence the critical value at α = 0.05 is t=2.131. Using these values, the confidence interval is 23±2.131*12/4 i.e. (16.607 <μ< 29.393)

b) Here n= 25, thus df = 24. Hence the critical value at α = 0.25 is t=1.15. Using these values, the confidence interval is 23±1.15*12/5 i.e. (20.24<μ<25.76)


a) H_0:μ=80 vs H_1:μ≠80, where μ is the average course in match course.

b) The test statistic is T=(68-80)/(11/√9)= -3.273. The test statistic follows a t-distribution with 8 degrees of freedom.

c) The p-value is given P(|T|>3.273)=2*P(T<-3.273) = 0.0113. It means that the probability of getting a sample that gives a similar or more extreme result than the obtained one is 0.0113.

d) As the p-value is less than the significance level of 0.05, we reject the null hypothesis. Thus, we conclude that there is sufficient evidence to support the claim that the average match score is different than