# Hypothesis test for a population mean

Monitoring the ecological health of the Everglades-the bottom temperatures are recorded at the Garfield Bight station and the mean of 19.7 degrees C is obtained for 62 temperatures on 62 different days. Assuming sigma=1.6 degrees C, test claim that the pop mean is less than 20.0 degrees C. Use a 0.05 significance level to calculate test statistic.

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Monitoring the ecological health of the Everglades-the bottom temperatures are recorded @ the Garfield Bight station and the mean of 19.7 degrees C is obtained for 62 temperatures on 62 different days. Assuming o=1.6 degrees C, test claim that the pop mean is less than 20.0 degrees C. Use a 0.05 significance level to calculate test statistic.

Solution:

This is a one tailed test of hypothesis using the Z or standard normal distribution as the reference distribution for the test statistic since the sample size is large and we are assuming the population standard deviation, , is known.

X is the random variable "temperature measurements". The sample size and sample mean are given in the problem, along with the population standard deviation.

Given in the problem

Null and alternative hypotheses

Note that the alternative hypothesis is "less than" since this is what we are interested in discovering, if it is true. In general, what you want to know or discover goes in the alternative hypothesis. The null hypothesis almost always has "=" in it.

Test statistic

If the null hypothesis is true, this test statistic is a randomly selected member of a standard normal distribution.

Rejection Region

At the level of significance, the critical valueof the rejection region is the 5th percentile (corresponding to ) of the Z, or standard normal distribution. From the table, this value is -1.645. The rejection region is therefore:

Test statistic < -1.645

Conclusion

Since the test statistic does not fall in the rejection region, at the 0.5 level of significance there is not enough evidence to reject the null hypothesis and conclude that the average temperature is less than 20.

P-value

Please note that an alternative to finding the rejection region, you could find the p-value and compare it to . If the p-value is less than , you can reject the null and conclude that the alternative is true. Otherwise there is insufficient evidence to reject the null. For this problem:

The p-value is greater than , giving us the same conclusion as with the rejection region method.

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