A random sample of 45 working mothers with six-year-old or younger children was

Question

A random sample of 45 working mothers with six-year-old or younger children was selected from companies that provide day-care facilities on premises and it was found that they missed an average of 6.4 days from work last year with a standard deviation of 1.2 days. Another sample of 50 such mothers were randomly selected from companies that do not provide day-care facilities on premises and it showed that these mothers missed an average of 9.3 days last year with a standard deviation of 1.85 days. Use a 0.05 significance level to test that the mean number of days missed per year by mothers working for companies that provide day-care facilities on premises is different than the mean number of days missed per year by mothers working for companies that do not provide day-care facilities on premises. what is the hypothesis? what is the t stat? and what is the p value? reject or retain?

Explanation / Answer

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   =   0  
Ha:   u1 - u2   =/   0   [HYPOTHESES]

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At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
Calculating the means of each group,              
              
X1 =    6.4          
X2 =    9.3          
              
Calculating the standard deviations of each group,              
              
s1 =    1.2          
s2 =    1.85          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    45          
n2 = sample size of group 2 =    50          
Thus, df = n1 + n2 - 2 =    93          
Also, sD =    0.31693848          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -9.150040732   [ANSWER, T STATISTIC]

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where uD = hypothesized difference =    0          
              
Also, using p values, by technology,              
              
p =    1.28563*10^-14       [ANSWER, P VALUE]

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As P < 0.05,    WE REJECT THE NULL HYPOTHESIS.          
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Hence, there is significant evidence that the mean number of days missed per year by mothers working for companies that provide day-care facilities on premises is different than the mean number of days missed per year by mothers working for companies that do not provide day-care facilities on premises. [CONCLUSION]

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