# We want to conduct a hypothesis test of the claim that the population mean score on a nationwide examination in anthropology is different from 492. So, we choose a random sample of exam scores. The sample has a mean of 505 and a standard deviation of 75. For each of the following sampling scenarios, choose an appropriate test statistic for our hypothesis test on the population mean. Then calculate that statistic. Round your answers to two decimal places. The sample has size 12, and it is from a normally distributed population with an unknown standard deviation. Oz = = 0 O It is unclear which test statistic to use.

We want to conduct a hypothesis test of the claim that the population mean score on a nationwide examination in anthropology is different from 492. So, we choose a random sample of exam scores. The sample has a mean of 505 and a standard deviation of 75. For each of the following sampling scenarios, choose an appropriate test statistic for our hypothesis test on the population mean. Then calculate that statistic. Round your answers to two decimal places. The sample has size 12, and it is from a normally distributed population with an unknown standard deviation. Oz = = 0 O It is unclear which test statistic to use.
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Marley Stone
In this hypothesis case, we need to test the claim that population mean score in anthropology is different from 492.
The t-test is used when the data is from normal distribution and the population standard deviation is unknown. In this case, as the population standard deviation is NOT KNOWN, hence we will use T-TEST.
Thus, the correct option is:
$\mu =\text{population mean}=492\phantom{\rule{0ex}{0ex}}\overline{x}=\text{sample mean}=505\phantom{\rule{0ex}{0ex}}s=\text{sample standard deviation}=75\phantom{\rule{0ex}{0ex}}n=\text{sample size}=12$
$t=\frac{\overline{x}-\mu }{s/\sqrt{n}}\phantom{\rule{0ex}{0ex}}=\frac{505-492}{75/\sqrt{12}}\phantom{\rule{0ex}{0ex}}\approx 0.60$
Hence, t=0.60