Question

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $88 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $2.50. Over the first 50 days she was employed at the restaurant, the mean daily amount of her tips was $89.93. At the 0.10 significance level, can Ms. Brigden conclude that her daily tips average more than $88?

1. State the null hypothesis and the alternate hypothesis.

• H₀: μ ≥ 88; H₁: μ < 88

• H₀: μ >88; H₁: μ = 88

• H₀: μ ≤ 88; H₁: μ > 88

• H₀: μ = 88; H₁: μ ≠ 88

2. State the decision rule.

• Reject H₀ if z > 1.28

• Reject H₁ if z < 1.28

• Reject H₁ if z > 1.28

• Reject H₀ if z < 1.28

3. Compute the value of the test statistic. (Round your answer to 2 decimal places.)

4. What is your decision regarding H₀?

• Reject H₀

• Do not reject H₀

5. What is the p-value? (Round your answer to 4 decimal places.)

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 88
Alternative hypothesis: u > 88

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n) S.E = 0.35355
b)

z_Critical = 1.28

c)

z = (x - u) / SE

z = 5.46

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

d) Interpret results. Since the z-value (5.46) lies in the rejection region, hence we have to reject the null hypothesis.

e)

The observed sample mean produced a t statistic test statistic of 5.46.

P-value = P(t > 5.46)

Use the t-value calculator for finding p-values.

P-value = 0.0000

Interpret results. Since the P-value (0.0000) is less than the significance level (0.10), we have to reject the null hypothesis.