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P1.T2.20.15. P-value and confidence intervals

Nicole Seaman

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Learning objectives: Explain what the p-value of a hypothesis test measures. Construct and apply confidence intervals for one-sided and two-sided hypothesis tests and interpret the results of hypothesis tests with a specific level of confidence. Identify the steps to test a hypothesis about the difference between two population means. Explain the problem of multiple testing and how it can bias results.

Questions:

20.15.1. Mary belies that the average net promoter score (NPS) in financial services, as a population, is at least 50. Her one-sided null hypothesis is H0: μ(NPS) ≤ 50 and her alternative hypothesis is HA: μ(NPS) > 50. Among a collected sample of 40 firms, her staff observes a sample average NPS of 53.60 with a standard deviation of 9.0. Her staff informs her that the test statistic is 2.53 and the two-sided p-value is 1.556% per the Excel function T.DIST.2T(2.530, 39) = 0.015563. Their report also includes these 95.0% critical t-values: T.INV(0.95, 39) = 1.685 and T.INV.2T(0.050, 39) = 2.023; as expected, these values are slightly higher than, respectively, the critical Z-values of 1.645 and 2.33. Each of the following is true EXCEPT which is false?

a. The power of Mary's one-sided test is 99.22%
b. The one-sided 95.0% confidence interval is [51.2, ∞)
c. Mary can reject a one-sided null hypothesis, H0: μ(NPS) ≤ 50, with 95.0% confidence
d. Mary can reject a one-sided null hypothesis, H0: μ(NPS) ≤ 50, with 99.0% confidence


20.15.2. Peter's boss asked him to collect the price to funds from operations (P/FFO) ratio for shopping center properties based on reported transactions. Peter was able to collect the P/FFO ratio for 49 shopping centers. The sample average P/FFO ratio is 11.30 and the standard deviation is 5.0. As usual in a realistic setting, neither Peter nor his boss knows the underlying population distribution from which the sample is drawn nor do they know the population variance.

Peter's one-sided null hypothesis is that the population's average P/FFO ratio is less than or equal to 10.0; i.e., his one-sided alternative hypothesis is HA: μ(P/FFO) > 10.0. For reference, in terms of the student's t distribution, the one- and two-sided critical t-values at 95.0% confidence are given by T.INV(0.95, 48) = 1.677 and =T.INV.2T(0.050, 48) = 2.011. Peter is discussing with his colleagues how to interpret the test.

I. Alice says the test statistic is 1.820
II. Bob says that Peter should reject with 95.0% confidence the one-sided null hypothesis, H0: μ(P/FFO) ≤ 10.0, because 1.820 > 1.677
III. Chris asserts that if Peter rejects the one-sided null, it is logical he can also reject with 95.0% confidence the two-sided null hypothesis, H0: μ(P/FFO) = 10.0
IV. Debra assert that Peter is incorrect to refer to t-statistics from the student's T distribution when he should use critical Z values from the normal distribution especially because this changes the decision
V. Eric says that we cannot use the confidence interval approach for this one-sided test because confidence intervals must have two bounds (upper and lower)

Who is (are) correct?

a. None of them are correct
b. Only Alice and Bob are correct; only I. and II. are true
c. Only Chris, Debra, and Eric are correct; only III. , IV. and V. are true
d. All of them are correct


20.15.3. A group of 36 borrowers agreed to test a firm's new product. The product aims to help the customer improve their credit score. Here are the before-versus-after results:
  • Before: average μ(b) = 620.0 and standard deviation σ(b) = 24.0
  • After: average μ(a) = 630.0 and standard deviation σ(a) = 30.0
According to these results, the sample mean improved by 10 points, although the dispersion also increased. The two groups are obviously not independent; it is assumed that their correlation is, ρ(a,b) = 0.70. We are treating these averages as sample means. The null hypothesis is that the population means are identical; i.e., H0: μ(b) = μ(a). With 95.0% two-tailed confidence, should we reject the null hypothesis and determine there has been an improvement in the average credit score?

a. No, we accept the null with 95.0% confidence because the test statistic is only 0.77
b. No, we accept the null with 95.0% confidence because the test statistic is only 1.56
c. Yes, we reject the null with 95.0% confidence and decide the sample mean has changed because the test statistic is 2.77
d. Yes, we reject the null with 95.0% confidence and decide the sample mean has changed because the test statistic is 5.38

Answers here:
 
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