Listed below are the numbers of words spoken in a day by each member of eight di

Question

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 15,998 25,634 1381 7482 19,389 15,252 14,392 25,616 Male Female 24,096 13,232 17,849 17,506 12,653 17,511 15,855 18,110 a. Use a 0.05 significance level to test the claim that among couples, males speak fewer words in a day than females ifferences d for the population of all pairs of data, where each individual difference d is defined as the words spoken In this example, d is the mean value of the d by the male minus words spoken by the female. What are the null and alternative hypotheses for the hypothesis test? Ho: Hal 11 word(s) H1 : Hal word(s) (Type integers or decimals. Do not round.) ldentify the test statistic. t(Round to two decimal places as needed.) ldentify the P-value P-value (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? Since the P-value is males speak fewer words in a day than females b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same the significance level the null hypothesis. There sufficient evidence to support the claim that conclusion reached in part (a)? The confidence interval is l word(s)

Explanation / Answer

Given that,
null, H0: Ud > 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.895
since our test is left-tailed
reject Ho, if to < -1.895
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -1458.5
We have d = -1458.5
pooled variance = calculate value of Sd= S^2 = sqrt [ 700019990-(-11668^2/8 ] / 7 = 9877.84
to = d/ (S/n) = -0.42
critical Value
the value of |t | with n-1 = 7 d.f is 1.895
we got |t o| = 0.42 & |t | =1.895
make Decision
hence Value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.4176 ) = 0.34437
hence value of p0.05 < 0.34437,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: -0.42
critical value: reject Ho, if to < -1.895
decision: Do not Reject Ho
p-value: 0.344

greater than , do not reject, insuffcient

Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = di/n
Sd = Sqrt( di^2 – ( di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( di/n ) =-11668/8=-1458.5
Pooled Sd( Sd )= Sqrt [ 700019990- (-11668^2/8 ] / 7 = 9877.841
Confidence Interval = [ -1458.5 ± t a/2 ( 5702.974/ Sqrt ( 8) ) ]
= [ -1458.5 - 2.365 * (3492.344) , -1458.5 + 2.365 * (3492.344) ]
= [ -9717.9 , 6800.9 ]

X Y X-Y (X-Y)^2 15998 24096 -8098 65577604 25634 13232 12402 153809604 1381 17849 -16468 271195024 7482 17506 -10024 100480576 19389 12653 6736 45373696 15252 17511 -2259 5103081 14392 15855 -1463 2140369 25616 18110 7506 56340036 -11668 700019990

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