Smoking during pregnancy is a known cause of reduced infant birth weight and oth
ID: 3262225 • Letter: S
Question
Smoking during pregnancy is a known cause of reduced infant birth weight and other issues that can affect the delivery and mortality. Despite this strong evidence, women who smoke find it difficult to quit even when they are pregnant. Suppose a study is done measuring the birth weight of babies of mothers who smoked during pregnancy and those who did not smoke. Use an alpha of 0.01. Use a non-parametric test. The birth weight of data in pounds are:
Smokers: 3.1, 4.2, 4.5, 5.0, 6.4, 4.7, 6.0
Nonsmokers: 5.5, 6.5, 7.1, 8.0, 6.8, 7.5, 6.2
a. State the null and alternative hypothesis.
b. Put this data into SPSS, and find the test statistic.
c. State the p-value and interpret. What does the p-value tell us in terms of the null hypothesis?
d. What can we conclude?
Explanation / Answer
Solution:-
Given,
Smokers: 3.1, 4.2, 4.5, 5.0, 6.4, 4.7, 6.0
Mean of smokers = 4.842857
Standard deviation of smokers = 1.1088
Nonsmokers: 5.5, 6.5, 7.1, 8.0, 6.8, 7.5, 6.2
Mean of non smokers = 6.8
Standard deviation of non smokers = 0.832666
Sample size, n1 = n2 = 7
a). State the null and alternative hypothesis.
Null hypothesis: 1 - 2 = 0 (i.e., there is no difference in the mean weights)
Alternative hypothesis: 1 - 2 0 (i.e., there is difference in the mean weights)
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
b). Put this data into SPSS, and find the test statistic.
Using SPSS, test statistics, t = -3.7343, or we can find this manually as follow,
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(1.10882/7) + (0.8326662/7] = 0.52410060478
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (1.10882/7 + 0.8326662/7)2 / { [ (1.10882 /7)2 / (6) ] + [ (0.8326662 / 7)2 / (6) ] }
DF = 0.07544989564 / (0.0051412123 + 0.00163506866) = 11.134
t = [ (x1 - x2) - d ] / SE = [ ( 4.842857 - 6.8) - 0 ] / 0.52410060478 = -3.7343
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
c). State the p-value and interpret. What does the p-value tell us in terms of the null hypothesis?
Since we have a two-tailed test, the P-value is the probability that a t statistic having 11 degrees of freedom is more extreme than -3.7343; that is, less than -3.7343 or greater than 3.7343.
We use the t Distribution Calculator to find P(t < -3.7343)
The P-Value is 0.0033.
The result is significant at p < 0.01
Interpret results. Since the P-value (0.0033) is less than the significance level (0.01), we cannot accept the null hypothesis.
d). What can we conclude?
Conclusion - Reject null hypothesis. We have significant evidence to prove the claim that the mean birth weight of babies of mothers who smoked during pregnancy and those who did not smoke is different.
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