Question

You carry out independent simple random samples of laptops produced by manufacturers Macrohard. Of the 25 Marcohard laptops sampled, 5 were defective. Of the pear laptops samples, 12 were defective. Clearly, more of the sampled pear laptops were defective than Macrohard; however, is this enough evidence to conclude the proportion of defective laptops is larger for Pear than Macrohard? To answer this perform an appropriate hypothesis test significance level a= 0.05.

1. check the assumptions for the hypothesis test

2. give the null and alternative hypothesis

3. calculate the test statistic

4. use either p-value or critical value to conclude

5. conclude the hypothesis test in context

Answer 1

Let the Macrohard sample be 1 and Pear laptop be 2.

For sample 1, we have that the sample size is N = 25, the number of favorable cases is X1 = 5, so then the sample proportion is ĝi = = = 0.2 For sample 2, we have that the sample size is N2 = 25, the number of favorable cases is X2 = 12, so then the sample proportion is ộ2 = * = e = 0.48 The value of the pooled proportion is computed as = $i=X3 = 35+23 5-12 = 0.34

1.

n₁*$\hat{p}_{1}$ = 5 > 5

n₁(1-$\hat{p}_{1}$) = 20 > 5

n₂* = 12 > 5

d

n₁(1-) = 13 > 5

d

i.e. all the assumptions are met for the test as $\hat{p}_{1}$ - follow normal distribution approximately.

d

2.

Ho: Pi = P2 Ha P1 P2

3.

2 P1 - P2 (1-7(1/nı +1/n2) 0.2 - 0.48 0.34. (1 - 0.34) (1/25 +1/25) -2.09

4.

Based on the information provided, the significance level is a = 0.05, and the critical value for a left- tailed test is ze = -1.64. The rejection region for this left-tailed test is R={z::< -1.64}

-2.09<Zc -1.64, it is then concluded that the null hypothesis is Since it is observed that z = rejected.

5.

Hence, we can say that proportion of defective laptops is larger for Pear than Macrohard

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