A random sample of eight students participated in a psychological test of depth

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A random sample of eight students participated in a psychological test of depth perce ption. 1. ged at a fixed distance apart at the far end of the laboratory. One by one the students were asked two judg e the distance between the two markers at the other end of the room. The sample data (in feet) were as follows: 2.1, 2,2, 2.6, 2.3, 1.8, 2.3, 2.4,2.5 At alpha 0.05 level of significance, test if the mean distance is more than 2 feet. Test statistict=srand Confidence Interval ±E withE= State the null and alternative hypothesis. State your conclusion for your hypothesis test.4 points Give the 95% Confidence Interval. Give a conclusion for the confidence interval test. 4 points points nts Hypothesis Test of 2Means 2. Do women candidates spend less in the campaigns for public office that men candidates? The following data represents the campaign expenditures of a randomly selected group of men and women candidates who have just completed their campaigns for public office Women: 132, 127, 134, 125, 127 Men: 134, 137, 135, 140, 130 At alpha 0.05 level of significance, test if women candidates spend less in the campaigns for public office that men candidates. Test the claim at 0.05 level of significance. Test statistic ni 4 points 4 points State the null and alternative hypothesis. State your conclusion for your hypothesls test

Explanation / Answer

Hypothesis
Given that,
population mean(u)=2
sample mean, x =2.15
standard deviation, s =0.487
number (n)=8
null, Ho: =2
alternate, H1: >2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.895
since our test is right-tailed
reject Ho, if to > 1.895
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =2.15-2/(0.487/sqrt(8))
to =0.8712
| to | =0.8712
critical value
the value of |t | with n-1 = 7 d.f is 1.895
we got |to| =0.8712 & | t | =1.895
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.8712 ) = 0.20627
hence value of p0.05 < 0.20627,here we do not reject Ho
ANSWERS
---------------
null, Ho: =2
alternate, H1: >2
test statistic: 0.8712
critical value: 1.895
decision: do not reject Ho
p-value: 0.20627

have evidence o support the claim

CONFIDENCE INTERVAL

TRADITIONAL METHOD
given that,
sample mean, x =2.15
standard deviation, s =0.487
sample size, n =8
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.487/ sqrt ( 8) )
= 0.172
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 2.365
margin of error = 2.365 * 0.172
= 0.407
III.
CI = x ± margin of error
confidence interval = [ 2.15 ± 0.407 ]
= [ 1.743 , 2.557 ]
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DIRECT METHOD
given that,
sample mean, x =2.15
standard deviation, s =0.487
sample size, n =8
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 7 d.f is 2.365
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 2.15 ± t a/2 ( 0.487/ Sqrt ( 8) ]
= [ 2.15-(2.365 * 0.172) , 2.15+(2.365 * 0.172) ]
= [ 1.743 , 2.557 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 1.743 , 2.557 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

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