Urgent please help. Amanda surveyed 20 juniors and seniors at delamar high school to find the number of hours per week they spen
d working at part-time jobs. Her results are shown below.
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1 answer:
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Answer:
a) the test statistic z = 1.891
the null hypothesis accepted at 95% level of significance
b) the critical values of 95% level of significance is zα =1.96
c) 95% of confidence intervals are (0.523 ,0.596)
Step-by-step explanation:
A survey of 700 adults from a certain region
Given sample sizes
Proportion of mean
<u>Null hypothesis H0</u> : assume that there is no significant difference between males and women reported they buy clothing from their mobile device
p1 = p2
<u>Alternative hypothesis H1:</u>- p1 ≠ p2
a) The test statistic is
where
on calculation we get p = 0.56
now q =1-p = 1-0.56=0.44
after calculation we get z = 1.891
b) The critical value at 95% confidence interval zα = 1.96 (from z-table)
The calculated z- value < the tabulated value
therefore the null hypothesis accepted
<u>conclusion</u>:-
assume that there is no significant difference between males and women reported they buy clothing from their mobile device
p1 = p2
c) <u>95% confidence intervals</u>
The confidence intervals are P± 1.96(√PQ/n)
we know that =
after calculation we get P = 0.56 and Q =1-P =0.44
Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))
now substitute values , we get
( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))
on simplification we get (0.523 ,0.596)
Therefore the population proportion (0.56) lies in between the 95% of <u>confidence intervals (0.523 ,0.596)</u>
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