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Independent random samples were selected from two binomial populations. The sizes and number of observed successes for each sample are shown in the following table

c. Explain how to use the confidence interval in part b to test the given hypothesis.

Short Answer

Expert verified

c) The 95% confidence interval is \(\left( {0.033,\,0.327} \right)\).

We reject the null hypothesis.

Step by step solution

01

 Given Information

The numbers of sample sizes are 100 and 125.

The probabilities of success are 58 and 50.

The hypothesis are given by

\(\begin{aligned}{l}{H_0}:{p_1} - {p_2} &= 0\\{H_a}:{p_1} - {p_2} \ne 0\end{aligned}\)

02

Explanation

The confidence interval is

\(\begin{aligned}{c}CI &= \left( {{{\hat p}_1} - {{\hat p}_2}} \right) \pm {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \\ &= \left( {0.58 - 0.4} \right) \pm {z_{0.025}}\sqrt {\frac{{.58 \times .42}}{{100}} + \frac{{.4 \times .6}}{{125}}} \\ &= .18 \pm 2.2414 \times 0.066\\ &= .18 \pm 0.147\\ = \left( {.18 - .147,.18 + .147} \right)\\ &= \left( {0.033,\,0.327} \right)\end{aligned}\)

03

Explanation

A confidence interval \(\left( {0.033,\,0.327} \right)\) is an interval centered about the sample statistic \(\left( {{\bf{2}}.{\bf{686}}} \right)\) with width equal to twice the margin of error.

We have, the 95% confidence interval is \(\left( {0.033,\,0.327} \right)\)

We see that the confidence interval does not contain the value defined by null hypothesis, and then our sample result is different.

Therefore, we reject the null hypothesis.

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Let t0 be a particular value of t. Use Table III in Appendix D to find t0 values such that the following statements are true.

a.=P(-t0<t<t0).95wheredf=10b.P(tโ‰ค-t0ortโ‰ฅt0)wheredf=10c.P(tโ‰คt0)=.05wheredf=10d.P(tโ‰ค-t0ortโ‰ฅt0)=.10wheredf=20e.P(tโ‰ค-t0ortโ‰ฅt0)=.01wheredf=5

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