Question

Consider the following hypothesis test: H 0: 50 H a: > 50 A sample of 50 is used and the population standard deviation is 6. Use the critical value approach to state your conclusion for each of the following sample results. Use = .05. a. With = 52.5, what is the value of the test statistic (to 2 decimals)? 2.42 Can it be concluded that the population mean is greater than 50? Yes b. With = 51, what is the value of the test statistic (to 2 decimals)? .97 Can it be concluded that the population mean is greater than 50? No c. With = 51.8, what is the value of the test statistic (to 2 decimals)? 1.65 Can it be concluded that the population mean is greater than 50? Yes

Answer 1

Answer:

Step-by-step explanation:

From the question we are told that:

Null Hypothesis $H_0: 50$

Alternative Hypothesis $H_a: > 50$

Sample size $n=50$

standard deviation$\sigma=6$

Significance level $\alpha=0.05$

a)

Sample mean $\=x=52.5$

Generally the equation for Test statistics is mathematically given by

$t=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }$

$t=\frac{52.5-50}{\frac{6}{\sqrt{50} } }$

$t=2.95$

Therefore from table

$Critical\ value=1.645$

We conclude The value of test statistics is greater than critical value.

Therefore we Reject the Null hypothesis $H_0$

b)

Sample mean $\=x=51$

Generally the equation for Test statistics is mathematically given by

$t=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }$

$t=\frac{51-50}{\frac{6}{\sqrt{50} } }$

$t=1.18$

Therefore from table

$Critical\ value=1.645$

We conclude,The value of test statistics is less than critical value.

Therefore we Fail to Reject the Null hypothesis $H_0$

c)

Sample mean $\=x=51$

Generally the equation for Test statistics is mathematically given by

$t=\frac{\=x-\mu}{\frac{\sigma}{\sqrt{n} } }$

$t=\frac{51.8-50}{\frac{6}{\sqrt{50} } }$

$t=2.12$

Therefore from table

$Critical\ value=1.645$

We conclude The value of test statistics is greater than critical value.

Therefore we Fail to Reject the Null hypothesis $H_0$