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Math
Posted 10 months ago

A quality control manager takes an SRS of 500 phone batteries from factory A, where 5%5 \% of the batteries have a defect. The manager also takes an independent SRS of 250 batteries from factory BB, where 9%9 \% of the batteries have a defect. The manager will then look at the difference (B−A)(B-A) between the proportions of defective batteries in each sample.

What are the mean and standard deviation of the sampling distribution of the difference in sample proportions?

Choose 1 answer:
(A) μp^B−p^A=0.04\mu_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=0.04
σp^B−p^A=0.05(0.95)500+0.09(0.91)250
\sigma_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=\sqrt{\frac{0.05(0.95)}{500}+\frac{0.09(0.91)}{250}}

(B) μp^B−p^A=0.04\mu_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=0.04
σp^B−p^A=0.05(0.95)500−0.09(0.91)250
\sigma_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=\sqrt{\frac{0.05(0.95)}{500}-\frac{0.09(0.91)}{250}}

(C) μp^B−p^A=0.07\mu_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=0.07
σp^B−p^A=0.05(0.95)500+0.09(0.91)250
\sigma_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=\sqrt{\frac{0.05(0.95)}{500}+\frac{0.09(0.91)}{250}}

(D) μp^B−p^A=0.07\mu_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=0.07
σp^B−p^A=0.05(0.95)500−0.09(0.91)250
\sigma_{\hat{p}_{\mathrm{B}}-\hat{p}_{\mathrm{A}}}=\sqrt{\frac{0.05(0.95)}{500}-\frac{0.09(0.91)}{250}}
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Answer from Sia
Posted 10 months ago
Solution by Steps
step 1
Calculate the mean difference in sample proportions: μp^B−p^A=pB−pA\mu_{\hat{p}_{B}-\hat{p}_{A}} = p_B - p_A
step 2
Substitute the given proportions: μp^B−p^A=0.09−0.05\mu_{\hat{p}_{B}-\hat{p}_{A}} = 0.09 - 0.05
step 3
Compute the mean difference: μp^B−p^A=0.04\mu_{\hat{p}_{B}-\hat{p}_{A}} = 0.04
step 4
Calculate the standard deviation of the difference in sample proportions: σp^B−p^A=pA(1−pA)nA+pB(1−pB)nB\sigma_{\hat{p}_{B}-\hat{p}_{A}} = \sqrt{\frac{p_A(1-p_A)}{n_A} + \frac{p_B(1-p_B)}{n_B}}
step 5
Substitute the given values into the standard deviation formula: σp^B−p^A=0.05×0.95500+0.09×0.91250\sigma_{\hat{p}_{B}-\hat{p}_{A}} = \sqrt{\frac{0.05 \times 0.95}{500} + \frac{0.09 \times 0.91}{250}}
step 6
Compute the standard deviation using the Asksia-LL calculator result: σp^B−p^A≈0.0205572\sigma_{\hat{p}_{B}-\hat{p}_{A}} \approx 0.0205572
[1] Answer
The mean and standard deviation of the sampling distribution of the difference in sample proportions are μp^B−p^A=0.04\mu_{\hat{p}_{B}-\hat{p}_{A}} = 0.04 and σp^B−p^A≈0.0205572\sigma_{\hat{p}_{B}-\hat{p}_{A}} \approx 0.0205572. The correct answer is (A).
Key Concept
Sampling Distribution of the Difference in Sample Proportions
Explanation
The mean of the sampling distribution of the difference in sample proportions is the difference between the two population proportions. The standard deviation is calculated using the formula that includes the proportions and sample sizes of both populations.

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