Question

Suppose the coffee industry claimed that the average adult drinks 1.7 cups of coffee per day. To test this​ claim, a random sample of 50 adults was​ selected, and their average coffee consumption was found to be 1.9 cups per day. Assume the standard deviation of daily coffee consumption per day is 0.5 cups using a significant level of 0.01

A. The​ z-test statistic is =

(round to two decimal places)

B. The critical z-score(s) are= (round to two decimal places)

(there should be two answers here one positive and one negative)

C. The p-value is=

(round to three decimal places)

Answer 1

Solution:

This is a two tailed test.

The null and alternative hypothesis is,

Ho: 1.7

Ha: 1.7

A)

The test statistics,

Z =( - )/ (/n)

= ( 1.9 - 1.7 ) / ( 0.5 / 50 )

= 2.83

B)

Critical value of  the significance level is α = 0.01, and the critical value for a two-tailed test is

= 2.58 and -2.58

C)

P-value = 2 * P(Z > z )

= 2 * ( 1 - P(Z < 2.83 ))

= 2 * 0.0023

= 0.0046

The p-value is p = 0.0046, and since p = 0.0046 < 0.01, it is concluded that the null hypothesis is rejected.

There is sufficient evidence to claim that the average adult drinks 1.7 cups of coffee per day.