Zvalue=μX−1500σX=1525−1500220/125=1.27Z0.025=−1.96Z0.975=1.96Zvalue∈(−1.96,1.96)Z_{value}= \frac {\mu_X-1500} {\sigma_X}= \frac {1525 - 1500} {220/ \sqrt{125}}= 1.27\\ Z_{0.025} = -1.96\\ Z_{0.975} = 1.96\\ Z_{value} \in (-1.96,1.96)Zvalue=σXμX−1500=220/1251525−1500=1.27Z0.025=−1.96Z0.975=1.96Zvalue∈(−1.96,1.96)
Hence, we cannot reject the null hypothesis that the population mean is equal to 1,500 at significant level of 0.05.
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