Summary statistics are given for independent simple random samples from two populations. Use the nonpooled t-test to conduct the required hypothesis test. 1 = 75.1, s 1 = 4.5, n 1 = 11, 2 = 66.2, s 2 = 5.1, n 2 = 9 Perform a two-tailed hypothesis test using a significance level of α = 0.01.

Respuesta :

Answer:

The decision rule is  

Reject the null hypothesis

Step-by-step explanation:

From the question we are told that

   The first sample mean is  [tex]\= x_1 = 75.1[/tex]

    The first sample standard deviation is  [tex]s_1 = 4.5[/tex]

    The first sample size [tex]n_1 = 11[/tex]

    The second sample mean is  [tex]\= x_2 = 66.2[/tex]

    The second sample standard deviation is  [tex]s_2 = 5.1[/tex]

     The second sample size is  [tex]n_2 = 9[/tex]

     The significance level is  [tex]\alpha = 0.01[/tex]

The null hypothesis is  [tex]H_o : \mu_1 = \mu_2[/tex]

The alternative hypothesis is  [tex]H_a : \mu_1 \ne \mu_2[/tex]

Generally the pooled standard deviation is mathematically represented as

      [tex]s = \sqrt{ \frac{(n_ 1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 -2 } }[/tex]

=>  [tex]s = \sqrt{ \frac{(11 - 1) 4.5^2 + (9 - 1) 5.1^2}{11 + 9 -2 } }[/tex]

=>  [tex]s = 4.78[/tex]

Generally the degree of freedom for the is mathematically represented as

       [tex]df = [ \frac{[\frac{s_1^2 }{n_1} + \frac{s_2^2}{ n_2} ]^2}{[\frac{[\frac{s_1^2}{n_1} ]^2}{n_1 - 1 } ] + [\frac{[\frac{s_2^2}{n_2} ]^2}{n_2 -1} ]} ][/tex]

=>   [tex]df = [ \frac{[\frac{4.5^2 }{11} + \frac{5.1^2}{9}  ]^2}{[\frac{[\frac{4.5^2}{11} ]^2}{ 11 - 1 } ] + [\frac{[\frac{5.1^2}{9} ]^2}{9 -1}  ]} ][/tex]

=>   [tex]df = 16 [/tex]

Generally the test statistics is mathematically represented as

      [tex]t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{s_1^2 }{n_1} + \frac{s_2^2}{n_2} } }[/tex]

=> [tex]t = \frac{ 75.1 - 66.2 }{ \sqrt{\frac{ 4.5^2 }{ 11} + \frac{5.1^2}{ 9} } }[/tex]

=> [tex]t = 4.092[/tex]

Generally the student distribution table the probability of  [tex]t = 4.092[/tex] to the right at a degree of freedom of  [tex]df = 16 [/tex]  is  

     [tex]P( T > 4.092 ) = 0.00042536[/tex]

Generally the p-value is mathematically represented as

      [tex]p-value = 2 * P(T > 4.092)[/tex]

=>    [tex]p-value = 2 * 0.00042536[/tex]

=>    [tex]p-value = 0.001[/tex]

From the value obtained we see that  [tex]p-value < \alpha[/tex] hence we

The decision rule is  

Reject the null hypothesis

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