In: Statistics and Probability
An investigator conducts 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 are collected and summarized below. 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. Make sure to 1. give the null and alternative hypotheses, 2. calculate the test statistic, and 3. give the conclusion (including a comparison to alpha or the critical value).
Group Number of children Mean PEF Std Dev PEF
Chronic Bronchitis 42 265 65
No Chronic Bronchitis 40 319 70
a) Hypotheses
HO: vs. HA
b) Calculate S2p=
c) Compute Test statistic T=
d) P-value
e) Conclusion
Accept H0 Reject H0
f) Calculate a 95% confidence interval for mean difference
Group | Number of children | Mean PEF | Std Dev PEF |
Chronic Bronchitis | 42 | 265 | 65 |
No Chronic Bronchitis | 40 | 319 | 70 |
The provided sample means are shown below:
Also, the provided sample standard deviations are:
and the sample sizes are
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.
(2)
Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:
c)
Test Statistics
Since it is assumed that the population variances are equal, the t-statistic is computed as follows:
d)
The p-value is p = 0.0003
e)
Reject H0
Since it is observed that t=−3.622<tc=−1.664, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p=0.0003, and since p=0.0003<0.05, it is concluded that the null hypothesis is rejected.
f)
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