The data table below contains the amounts that a sample of nine customers spent
ID: 3304660 • Letter: T
Question
The data table below contains the amounts that a sample of nine customers spent for lunch (in dollars) at a fast-food restaurant Complete parts (a) through (d). Click here to view page 1 of the table of critical values of t 4.18 5.04 5.89 6.36 7.48 7.62 8.58 8.53 9.77 a. At the 0.01 level of significance, is there evidence that the mean amount spent for lunch is different from $6.50? State the null and alternative hypotheses H1 : (Type integers or decimals. Do not include the $ symbol in your answer.) Identify the critical value(s). The critical value(s) is(are) (Round to four decimal places as needed. Use a comma to separate answers as needed.) Determine the test statistic The test statistic is (Round to four decimal places as needed.) State the conclusion. Ho- There is evidence to conclude that the mean amount spent for hnch is different from $6 50 b. Determine the p-value and interpret its meaning. The p-value is (Round to four decimal places as needed.) Interpret the meaning of the p-value. Choose the correct answer below A. The p-value is the probability of obtaining a sample mean that is equal to or more extreme than SO 55 away from $6.50 if the null hypothesis is toe. O B. The p-value is the probability of obtaining a sample mean that is equal to or more extreme than S0.55 below $6.50 if the null hypothesis is false. C The p-value is the probability of obtaining a sample mean that is equal to or more extreme than $0 55 above $0.50 if the null hypothesis is false. O D. The p-value is the probability of not rejecting the null hypothesis when it is false. c. What assumption must you make about the population distribution in order to conduct the t test in (a) and (b)? O A. The population distribution of the amount spent is normally distributed O B. The population distribution of the amount spent follows the Students t distribution. O C. The population distribution of the amount spent is skewed to one side.Explanation / Answer
The mean of the data is 7.05 and the standard deviation is 1.82 , also the number of data points = 9
a) H0 : mu is not different from 6.5
h1 : mu is different from 6.5
to calculate the test stat we use the formula
(mu-mean)/(sd/(sqrt(n)))
(7.02-6.5)/(1.8/sqrt(9)) = 0.866
, now we check the t table to check the critical value for df = n-1 = 9-1 = 8
the critical value is 2.33
as the t stat < t critical hence we fail to reject the null hypothesis and conclude that the mean value is not different from 6.5
we check from the t table that p value is close to 0.411
p pvalue is the probability of not rejecting the null hypothesis when it is false
in order to carry out the test we must assume that the data follows a t distribution
when the sample size is small , it is difficult to evaluate the t distribution assumption because we have too few data points to check the shape of the distribution. we can check the mean and median are close in value
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