A well-known brokerage firm executive claimed that 60% of
investors are currently confident of meeting their investment
goals. An XYZ Investor Optimism Survey, conducted over a two week
period, found that in a sample of 200 people, 65% of them said they
are confident of meeting their goals.
Test the claim that the proportion of people who are confident is
larger than 60% at the 0.10 significance level.
The null and alternative hypothesis would be:
H0:μ=0.6H0:μ=0.6
H1:μ≠0.6H1:μ≠0.6
H0:μ=0.6H0:μ=0.6
H1:μ<0.6H1:μ<0.6
H0:μ=0.6H0:μ=0.6
H1:μ>0.6H1:μ>0.6
H0:p=0.6H0:p=0.6
H1:p≠0.6H1:p≠0.6
H0:p=0.6H0:p=0.6
H1:p>0.6H1:p>0.6
H0:p=0.6H0:p=0.6
H1:p<0.6H1:p<0.6
The test is:
two-tailed
left-tailed
right-tailed
The test statistic is: (to 3 decimals)
The p-value is: (to 4 decimals)
Based on this we:
H0: p = 0.60
Ha: p > 0.60
The test is right tailed
Test statistics
z = ( - p) / sqrt [ p( 1 - p) / n ]
= (0.65 - 0.60) / sqrt ( 0.60 * ( 1 - 0.60) / 200 ]
= 1.443
p-value = P(Z > z)
= P(Z > 1.443)
= 1 - P(Z < 1.443)
= 1 - 0.9255
= 0.0745
Since p-value < 0.10 significance level, Reject the null hypothesis.
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