Question

Harry’s Hamburgers claims that the residents of Harryville eat at the hamburger chain an average of exactly 18 times per year. You took a sample of 48 residents of Harryville and found that the sample mean was 17. The sample standard deviation was 4.5. You want to test to see if Harry’s claim can be disputed with a significance level of 0.05.

1. For each number in the problem, give the appropriate mathematical symbol associated with that number:

a. 18 = b. 48 = c. 17 = d. 4.5 = e. 0.05 =

2. State the null and alternative hypotheses.

3. How many tails is this test?

4. Specify α.

5. Give the symbol for the test statistic and compute the value.

6. How many degrees of freedom are there?

7. Use the t table to find a range for the p-value.

8. What is your conclusion using the p-value approach in statistical terms?

9. What is your conclusion in business terms?

10. Now do the problem using the confidence interval approach and state your conclusion in statistical terms.

Answer 1

1) a) $\mu$ = 18

b) n = 48

c) $\bar x$ = 17

d) s = 4.5

e) $\alpha$ = 0.05

2)

Ho :μ = 18

  

H, :μ+ 18

3) This is a two-tailed test.

4) $\alpha$ = 0.05

5) The test statistic is t

- s/n

17 – 18 4.5/48

  

= -1.54

6) DF = 48 - 1 = 47

7) P-value = 2 * P(T < -1.54)

= 2 * 0.0651

= 0.1302

0.1 < P-value < 0.2

8) Since the P-value is greater than $\alpha$ , so we should not reject H₀.

9) At 0.05 significance level, there is sufficient evidence to support Harry's claim. So Harry's claim cannot be disputed.

10) At 95% confidence level, the critical value is t^* = 2.012

The 95% confidence interval is

.. * ¥于C

4.5 = 17 + 2.012 * V48

= 17+1.3068

= 15.6932, 18.3068

Since the interval contains the hypothesized value 18, so we should nor reject the null hypothesis.