The TI-83/84 Plus calculator can be used to generate random data from a normally

Question

The TI-83/84 Plus calculator can be used to generate random data from a normally distributed population. The command randNorm (74, 12.5, 100) generates 100 values from a normally distributed population with u=74 or      (sigma)   =12.5 (for pulse rates of women.) One such generated sample of 100 values has a mean of 74.4 and a standard deviation of 11.7. Assume that     (lpha)    is known to be 12.5 and use a .05 significance level to test the claim that the sample actually does come from a population with a mean equal to 74. Use the p-value method. Based on the results, does it appear that the caalculator's random number generator is working correctly? not using a graphing calculator.

Explanation / Answer

Solution:

Here, we have to use the one sample z test for the population mean. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: The sample comes from a population with a mean equal to 74.

Alternative hypothesis: Ha: The sample does not come from a population with a mean equal to 74.

In symbolic notation,

Null hypothesis: H0: µ = 74 versus Alternative hypothesis: Ha: µ 74

This is a two tailed test.

Here, we are given

Population mean = µ = 74

Population standard deviation = = 12.5

Sample size = n = 100

Sample mean = Xbar = 74.4

Sample standard deviation = S = 11.7

Level of significance = = 0.05

The test statistic formula for this test is given as below:

Test statistic = Z = (Xbar - µ)/[/sqrt(n)]

Test statistic = Z = (74.4 – 74.0) / [12.5/sqrt(100)]

Test statistic = Z = 0.4/[12.5/10] = 0.4/1.25 =0.32

Test statistic = Z = 0.32

Critical values = -1.96 and 1.96

P-value = 0.7490

= 0.05

P-value >

So, we do not reject the null hypothesis that population mean is 74.

This means we conclude that there is sufficient evidence that the sample actually come from a population with a mean equal to 74.

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