The following display from a TI-84 calculator presents the results of a hypothes

Question

The following display from a TI-84 calculator presents the results of a hypothesis test for a population mean mu. mu =/2.0 t = -2.424366107 p =.013385940 x = 1.88 S_x =.21 n = 18 State the null and alternative hypotheses. How many degrees of freedom are there? Do you reject H_0 at alpha =.05? Explain. Someone want to test the hypothesis H_0: mu = 1.8 versus H_a: mu > 1.8. Use the information in the original display above to compute the t-statistic for this test. Compute the P-value for the test described in part (d). Can the null hypothesis in part (d) be rejected at alpha =.05? Explain.

Explanation / Answer

a)H0: mu =2

Ha: mu not equals 2

Two tailed test.

b)Deg of freedom = n-1 = 18-1 = 17

c)p =0.013 <0.05

Hence reject H0

d) For one tailed hypothesis with mu = 1.8

t = 0

p =0.5

Since p>0.05, accept H0

e) p = 0.5

f) No cannot be rejected.

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