Please help me to get question D, the p value (1 point) The following Exercise i
ID: 3374590 • Letter: P
Question
Please help me to get question D, the p value
(1 point) The following Exercise is based on summary statistics rather than raw data. This information is typically all that is presented in published reports. You can calculate inference procedures by hand from the summaries. Use the conservative Option 2 (degrees of freedom the smaller of n1-1 and n2 - 1 for two-sample t confidence intervals and P-values. You must trust that the authors understood the conditions for inference and verified that they apply. This isn't always true.) Equip male and female students with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Count the words each subject speaks during each recording period, and from this, estimate how many words per day each subject speaks. The published report includes a table summarizing six such studies. Here are two of the six: Estimated Average Number (SD) of Words Spoken per Day Sample Size Study Women Men Women Men 15353 15680 (7819) 16533 (9067) 2 29 23 16008 (7584) 12462 (8179) Readers are expected to understand this to mean, for example, the 53 women in the first study had x = 15680 and S-7819 It is commonly thought that women talk more than men. Does either of the two samples support this idea? For each study: (a) State the alternative hypothesis in terms of the population mean number of words spoken per day for men (HM) and for women (u). If necessary use !- to represent ; Study 1: HMExplanation / Answer
Solution:-
d)
Study 1
P-value(0.697) is greater than 0.05.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: uM> uF
Alternative hypothesis: uM < uF
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 1644.59
DF = 52
t = [ (x1 - x2) - d ] / SE
t = - 0.519
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize 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 population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of - 0.519.
Therefore, the P-value in this analysis is 0.692
Interpret results. Since the P-value (0.697) is greater than the significance level (0.05), we cannot reject the null hypothesis.
Study 2
0.05 < p-value < 0.10
P-value(0.692) is greater than 0.05.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: uM> uF
Alternative hypothesis: uM < uF
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 2211.7573
DF = 22
t = [ (x1 - x2) - d ] / SE
t = 1.60
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize 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 population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of 1.60.
Therefore, the P-value in this analysis is 0.062
Interpret results. Since the P-value (0.062) is greater than the significance level (0.05), we cannot reject the null hypothesis.
0.05 < p-value < 0.10