Question

Starbucks serves several sizes of coffee. The Grande is 16 fluid oz. There is some variation from cup to cup because the filling machinery is not perfectly precise. Data on past production shows that the distribution of the contents is close to Normal, with standard deviation σ = 0.5 oz. To assess the state of the filling process at a given Starbucks store, 70 cups were randomly selected over the course of a week. The sample mean content is found to be 15.85 oz. Is this value convincing evidence that the mean fill of all Grande cups of coffee produced by the current process differs from the desired level of 16.0 oz.?

The hypotheses: H₀: μ = 16.0 and H_A: μ ≠ 16.0. The estimate is the sample mean.

Calculate P-value of the test. Note: you may need to use Excel to perform this calculation.

Group of answer choices

0.0500

0.0316

0.0158

0.0121

0.0060

Answer 1

The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case,
the p-value is p =P(|Z|>2.51)=0.0121
Using excel formula :

0.0121

=(1-NORM.S.DIST(ABS(-2.51),TRUE))*2

Entire test:

One-Sample Z test

The sample mean is Xˉ=15.85, the population standard deviation is σ=0.5, and the sample size is n=70. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =16 Ha: μ ≠16 This corresponds to a Two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Two-tailed test is Zc​=1.96. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Two-tailed test is |Z|>1.96 i.e. Z>1.96 or Z<-1.96 (3) Test Statistics The z-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(|Z|>2.51)=0.0121 Using excel formula : 0.0121 =(1-NORM.S.DIST(ABS(-2.51),TRUE))*2 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |Z|=2.51 > Zc​=1.96, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0121, and since p=0.0121≤0.05, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ  is different than 16, at the 0.05 significance level.

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