Question

1. given the minitab output:

test and ci for one proportion
test of p = 0.5 vs p ≠ 0.5
sample
x n sample p
1 175 200 0.875000 (0.803233, 0.928182)
99% ci
z* = 2.576
p - value
0.0000
fill in the blanks:
a. ho: p = 0.05 ha: p ≠ 0.5
b. type of test: z test: srs 15/15
c. α = 0.01
d. compute the z - value: __
e. decision: __
f. conclusion: there is

Explanation:

Step1: Hypothesis Setup

The null hypothesis \(H_0:p = 0.05\) and the alternative hypothesis \(H_a:p
eq0.05\) are given. This is a two - tailed test.

Step2: Significance Level

The significance level \(\alpha=0.01\) is provided.

Step3: Compute Z - value

The formula for the test statistic \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(\hat{p}=\frac{x}{n}=\frac{175}{200}=0.875\), \(p = 0.5\), and \(n = 200\).

$$
LATEXBLOCK0
$$

Step4: Decision

Since the p - value \(=0.0000\lt\alpha = 0.01\), we reject the null hypothesis.

Step5: Conclusion

We conclude that there is sufficient evidence to support the claim that \(p
eq0.5\) at the \(\alpha = 0.01\) significance level.

Answer:

a. \(H_0:p = 0.05\), \(H_a:p
eq0.5\)
b. Two - tailed \(z\) test
c. \(\alpha=0.01\)
d. \(z\approx10.6\)
e. Reject \(H_0\)
f. There is sufficient evidence to support \(p
eq0.5\) at \(\alpha = 0.01\)