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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

User Kask
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6 votes

Answer:

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=(\=x-\mu)/((\sigma)/(√(n) ) )


t=(52.5-50)/((6)/(√(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=(\=x-\mu)/((\sigma)/(√(n) ) )


t=(51-50)/((6)/(√(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=(\=x-\mu)/((\sigma)/(√(n) ) )


t=(51.8-50)/((6)/(√(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

User Mithun Rana
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