We know that if pn 5 and 1 - pn 5 we can assume sample


Question 1: Internet upload speed The technical operations department wants to ensure that the mean target upload speed for all Internet service subscribers is at least 0.97 on a standard scale in which the target value is 1.0. Each day, upload speed was measured 50 times, with the following results: 0.854 1.023 1.005 1.030 1.219 0.977 1.044 0.778 1.122 1.114 1.091 1.086 1.141 0.931 0.723 0.934 1.060 1.047 0.800 0.889 1.012 0.695 0.869 0.734 1.131 0.993 0.762 0.814 1.108 0.805 1.223 1.024 0.884 0.799 0.870 0.898 0.621 0.818 1.113 1.286 1.052 0.678 1.162 0.808 1.012 0.859 0.951 1.112 1.003 0.972 Compute the sample statistics and determine whether there is evidence that the population mean upload speed is less than 0.97.

Question 2: (Hypothesis test for the proportion on small sample size) We know that if πn ≥ 5 and (1 - π)n ≥ 5, we can assume sample distribution is normal because of central limit theorem. Then we can use Z-test to do the hypothesis testing. But if the sample size is small, we cannot assume the sample distribution is normal. Here is the question: One manufacturer's high-quality product rate is 30%. Recently, a sample with size 15 products is collected. Among them, 3 are with high quality. Can we say the high-quality rate remains 30% using level of significance α = 0.05? We can find that nπ = 15 × 0.3 = 4.5 ≤ 5(In reality, nπ is better be much larger than 5), so we do not take the assumption that it is normally distributed. We cannot use Z-test. This hypothesis is still two-tail hypothesis and the rejection region is {x ≤ c1 or x ≥ c2}. c1 and c2 are two critical values. Can you find c1 and c2 to get P-value and do the hypothesis testing? (Hint: binomial distribution)

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Finance Basics: We know that if pn 5 and 1 - pn 5 we can assume sample
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