Question

In: Math

Researchers conducted a study to investigate whether there is a difference in mean PEF in children...

Researchers conducted a study to investigate whether there is a difference in mean PEF in children with chronic bronchitis as compared to those without. Data on PEF were collected from 100 children with chronic bronchitis and 100 children without chronic bronchitis. The mean PEF for children with chronic bronchitis was 290 with a standard deviation of 64, while the mean PEF for children without chronic bronchitis was 308 with a standard deviation of 77. Based on the data, is there statistical evidence of a lower mean PEF in children with chronic bronchitis as compared to those without? Run the appropriate test at α=0.05. Assume equal variances. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available.

Group

Number of Children

Mean PEF

Std Dev PEF

Chronic Bronchitis

100

290

64

No Chronic Bronchitis

100

308

77

Solutions

Expert Solution

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

( 2 ) T-test for two Means – Unknown Population Standard Deviations ( equal variances )

( 3 ) Decision rule :

Based on the information provided, the significance level is alpha = 0.05α=0.05, and the degrees of freedom are d.f=198. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is t_c = 1.972  for α=0.05 and d f = 198

The rejection region for this two-tailed test is R = { t : ∣t∣ > 1.972 }.

( 4 ) Test statistic :

( 5 ) Since it is observed that |t| = 1.798 ≤ t c​=1.972, it is then concluded that the null hypothesis is not rejected.


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