The owner of a local nightclub has recently surveyed a random sample of n=250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. A random sample was taken. The sample mean was 30.45 and the sample standard deviation was 5 years. Using the 0.10 level of significance, is there evidence that the mean age of her customer is over 30? a.) State the null hypothesisb.) State the alternative hypothesisc.) What critical value should be used? d.) What is the test statistic? e.) What is the p-value? f.) Interpret the meaning of the p-value? g.) Should the null hypothesis be rejected? h.) What is the conclusion? i.) State one of the assumptions needed? j.) State the degree of freedom

Question

Mathematics

Oompa Loompa

15 June, 14:01

The owner of a local nightclub has recently surveyed a random sample of n=250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. A random sample was taken. The sample mean was 30.45 and the sample standard deviation was 5 years. Using the 0.10 level of significance, is there evidence that the mean age of her customer is over 30? a.) State the null hypothesisb.) State the alternative hypothesisc.) What critical value should be used? d.) What is the test statistic? e.) What is the p-value? f.) Interpret the meaning of the p-value? g.) Should the null hypothesis be rejected? h.) What is the conclusion? i.) State one of the assumptions needed? j.) State the degree of freedom

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Answers (1)

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Brielle Duffy · Answer 1

Step-by-step explanation: We would set up the hypothesis test. a) For the null hypothesis,

µ ≤ 30

b) For the alternative hypothesis,

µ > 30

It is a right tailed test.

c) the critical value is the t test statistic value

d) Since no population standard deviation is given, the distribution is a student's t.

Since n = 250,

Degrees of freedom, df = n - 1 = 250 - 1 = 249

t = (x - µ) / (s/√n)

Where

x = sample mean = 30.45

µ = population mean = 30

s = samples standard deviation = 5

t = (30.45 - 30) / (5/√250) = 1.42

e) We would determine the p value using the t test calculator. It becomes

p = 0.078

f) Since alpha, 0.1 is > than the p value, 0.078, then we would reject the null hypothesis.

g) the population and the data sampled is normally distributed

f) degree of freedom = 249