Question

5. Consider running a test on the proportion of students who are doing regular physical activity. You are trying to disprove the claim that 30% of students do regular physical activity, and therefore choose the following hypotheses Ho:p=0.3 H, : p 0.3 and plan to use a 5% significance level. Calculate the prob- ability of committing a type II error when using a sample of n 50 students, assuming the true proportion is 0.4. In other words, find β(0.4).

Answer 1

Let $\hat{p}$ denote the sample proportion of students who are doing a regular physical activity.

We are given that n = 50

Now, Under H₀:

P-P p(1-p 7n

Our test statistic is :

$\begin{align*} z &= \frac{\hat{p}-0.3}{\sqrt{\frac{0.3(1-0.3)}{50}}} \\ &= \frac{\hat{p}-0.3}{0.064807} \end{align*}$

Now, since it is a two tailed test and the significance level is 5%, we reject H₀ if the absolute value of the test statistic is greater z_0.05/2 = z_0.025 = 1.959964.

$\text{Reject H0 if }\left |{\frac{\hat{p}-0.3}{0.064807}} \right | > 1.959964 \\ \Rightarrow \text{Reject H0 if } \left |\hat{p}-0.3 \right | > 0.127020 \\ \Rightarrow \text{Reject H0 if } (\hat{p}-0.3) > 0.127020 \ \ or \ \ (\hat{p}-0.3) <-0.127020 \\ \Rightarrow \text{Reject H0 if } \hat{p} > 0.427020 \ \ or \ \ \hat{p} < 0.172980$

+ Do not reject HO if 0.172980 < p < 0.427020

Now, we can calculate the type II error.

$\begin{align*} \beta(0.4) &= P(\text{Not rejecting H0 } | \ p = 0.4) \\ &= P(0.172980 < \hat{p} < 0.427020 \ | \ p = 0.4) \\ &= P\begin{pmatrix} \frac{0.172980 - 0.4}{\sqrt{\frac{0.4(1-0.4)}{50}}} < \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} < \frac{0.427020 - 0.4}{\sqrt{\frac{0.4(1-0.4)}{50}}} \end{pmatrix} \\ &= P(-3.27675 < N(0,1) < 0.39000) \\ &= P(N(0,1)<0.39000) - P(N(0,1) \le -3.27675) \\ &= 0.651733 - 0.000525 \\ &= \bf{0.651208 \ \ \ \ [Answer]} \end{align*}$

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