Since the P-value is less than or equal to, or greater than the siginificance level, reject or warrent rejection of the null hypothesis. There is or is not suffcient evidence to support the claim that males speak fewer words in a day than females. Since the confidence interval contains zero, only positive values, or only negative values | reject or fail to reject the null hypothesis. **Word Counts in Couples Study**The table below presents the number of words spoken in a day by each member of eight different randomly selected couples. The focus is on comparing word usage between males and females. The data is as follows:- **Male Word Count**: 16,124; 25,956; 1,427; 7,739; 18,900; 15,594; 14,356; 26,609- **Female Word Count**: 25,069; 13,852; 18,788; 18,151; 13,262; 16,496; 16,415; 18,857**Hypothesis Testing**We will use a significance level of 0.01 to evaluate the claim that males speak fewer words in a day than females. Here, \(\mu_d\) represents the mean value of the differences, \(d\), for all pairs of data, where each individual difference, \(d\), is defined as:\[ d = \text{words spoken by the male} - \text{words spoken by the female} \]**Null and Alternative Hypotheses**Let's define the null hypothesis, \(H_0\), and the alternative hypothesis, \(H_1\):- \(H_0: \mu_d = 0\) (Males speak the same number of words as females)- \(H_1: \mu_d < 0\) (Males speak fewer words than females)Fill in the blanks with integers or decimals as necessary. Do not round the numbers. Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. Complete parts (a) and (b) below.| | Male | Female ||-------|--------|--------|| Pair 1| 16,124 | 25,069 || Pair 2| 25,956 | 13,852 || Pair 3| 1,427 | 18,788 || Pair 4| 7,739 | 18,151 || Pair 5| 18,900 | 13,262 || Pair 6| 15,594 | 16,496 || Pair 7| 14,356 | 16,415 || Pair 8| 26,609 | 18,857 |**a. Use a 0.01 significance level to test the claim that among couples, males speak fewer words in a day than females.**In this example, \( \mu_d \) is the mean value of the differences \( d \) for the population of all pairs of data, where each individual difference \( d \) is defined as the words spoken by the male minus words spoken by the female. What are the null and alternative hypotheses for the hypothesis test?\[ H_0 : \mu_d = 0 \] word(s) \[ H_1 : \mu_d < 0 \] word(s)(Type integers or decimals. Do not round.)

Given information: Male Female 16124 25069 25956 13852 1427 18788 7739 18151 18900 13262 15594 16496 14356 16415 26609 18857a) Null hypothesis: HA : μDifference=0 Alternate hypothesis: HA : μDifference<0 This is one tailed test. The significance level is 0.01 and we will conduct a matched-pairs t-test of the null hypothesis. Use the given data to form the excel table: S.No Male Female d d-d¯2 1 16124 25069 -8945 51435791.02 2 25956 13852 12104 192574598.3 3 1427 18788 -17361 242981847 4 7739 18151 -10412 74630161.27 5 18900 13262 5638 54924773.77 6 15594 16496 -902 758858.7656 7 14356 16415 -2059 81724.51563 8 26609 18857 7752 90728006.27 Sum 126705 140890 -14185 708115760.9 Mean 15838.125 17611.25 -1773.125 88514470.11Use the sample data to calculate the standard deviation of the differences(s), the standard error (S.E) of the mean difference, the degrees of freedom(DF), and the t statistic test statistic(t). s=∑(d-d¯)2n-1=708115760.98-1=10057.80266s=10057.80S.E=sn=10057.808=3555.97D.F=n-1=8-1=7t=[(x1-x2)-D]S.E=-1773.125-03555.97=-0.498633252=-0.50The P-value for this test is the probability that a t statistic having 7 degrees of freedom is less than -0.50. P-value = P(t < -0.50) Use the excel formula, "=T.DIST(-0.5,7,TRUE)" to determine the P-value. P-value = 0.3162 Decision: Since the P-value (0.3162) is greater than the significance level (0.01), hence we failed to reject the null hypothesis. Conclusion: Since the P-value is greater than the significance level, hence we failed to reject the null hypothesis. There is not sufficient evidence to support the claim that males speak fewer words in a day than females.