Question

A cooking article reported that cooking a quality meal takes time; in fact, the article states that it takes longer than 45.0 minutes. Suppose a study is conducted to test the validity of this statement at a 95.00% confidence interval/level. A sample of 18 people is selected, and the length of time to put a meal on the table is listed: 45.20 40.70 41.10 49.10 30.90 45.20 55.30 52.10 45.40 55.10 38.80 43.10 39.20 58.60 49.80 43.20 47.90 46.60 What will be the hypothesis in this study? What will be the area of rejection in this study using p critical value? Create the statement of rejection or not-rejection based upon the p value and the p critical value What will be the conclusion of this study?

Answer 1

the null ad alternative hypothesis for this problem is

H0:u =45

Ha:u>45 (one tailed)

using excel>addin>phstat>one sample t

we have

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	45
Level of Significance	0.05
Sample Size	18
Sample Mean	45.96111111
Sample Standard Deviation	6.814097747
Intermediate Calculations
Standard Error of the Mean	1.6061
Degrees of Freedom	17
t Test Statistic	0.5984
Upper-Tail Test
Upper Critical Value	1.7396
p-Value	0.2787
Do not reject the null hypothesis

Rejection region : we will reject the null hypothesis is p value of the test statistic will be greater than 0.05

the value of test statistic t = 0.5984

p value is 0.2787

since p value is greater than 0.05 so we do not reject Ho and conclude that cooking a quality meal takes time takes longer than 45.0 minutes