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City health officials will conduct a two-sample t-test for a difference in means to investigate whether
stray dogs in the city have more fleas, on average, than do stray cats. A random sample of 22 stray dogs
had a mean of 120 fleas and a standard deviation of 20 fleas. A random sample of 15 stray cats had a
mean of 100 fleas and a standard deviation of 30 fleas.
Assuming a null hypothesis of no difference, which of the following is a correct test statistic for the test?

Respuesta :

Answer:

The calculated value     t =   2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means      

Step-by-step explanation:

Step(i):-

Given that the first sample size 'n₁' = 22

Given that the mean of the first sample x₁⁻ = 120

Given that the standard deviation of the sample (s₁) = 20

Given that the mean of the second sample x₂⁻ = 100

Given that the standard deviation of the second sample (S₂) = 30

Step(ii):-

T-test statistic

               [tex]t = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }[/tex]

     where

             [tex]S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2} S_{2} ^{2} }{n_{1} +n_{2}-2 }[/tex]

            [tex]S^{2} = \frac{22( 20)^{2}+ 15(30) ^{2} }{22 +15-2 }[/tex]

           S² =  637.1428

           S = √637.1428 = 25.24168  

The standard deviation of the sample 'S' = 25.24168   

Step(iii):-

Null Hypothesis:H₀:

There is no difference between the means

Alternative Hypothesis:H₁:

There is a difference between the means

         T-test statistic

               [tex]t = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }[/tex]

              [tex]t = \frac{ 120 -100 }{\sqrt{637.14 (\frac{1}{22 }+\frac{1}{15 } ) } }[/tex]

             t =   2.366

Degrees of freedom

           γ = n₁+n₂ -2 = 22+15-2 = 35

     t₀.₀₅ = 2.0301

The calculated value     t =   2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means          

The correct test statistic for the test of the hypothesis for the considered situation is 2.2619 approximately.

How to perform two sample t-test?

If the sample sizes < 30, and we want to test the difference between the sample means, then we perform t-test.

Let we have:

  • [tex]\overline{x}_1[/tex] = mean of first sample
  • [tex]\overline{x}_2=[/tex] mean of second sample
  • [tex]s_1[/tex] = standard deviation of first sample
  • [tex]s_2[/tex] = standard deviation of second sample.
  • [tex]n_1[/tex] = first sample's size
  • [tex]n_2[/tex] = second sample's size

Then, the value of t-test statistic is obtained as:

[tex]t = \dfrac{\overline{x}_1 - \overline{x}_2}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}}[/tex]

If the level of significance is [tex]\alpha[/tex], then as we have:

the degree of freedom (d.f) = [tex]n_1 + n_2 - 2[/tex],

the critical value of t is found to be [tex]t_{\alpha/2}[/tex] , then if we get:

[tex]|t| < t_{\alpha/2}[/tex] we cannot reject the null hypothesis

and if we get [tex]|t| > t_{\alpha/2}[/tex] , then we may reject the null hypothesis and thus, accept the alternate hypothesis.

For this case, we're specified that:

For sample of stray dogs:

  • [tex]n_1 = 22[/tex]
  • [tex]\overline{x}_1 = 120[/tex]
  • [tex]s_1 = 20[/tex]

For sample of stray cats:

  • [tex]n_2= 15[/tex]
  • [tex]\overline{x}_2 = 100[/tex]
  • [tex]s_2 = 30[/tex]

Thus, the t-test statistic for the difference of mean of these two samples is calculated as:

[tex]t = \dfrac{\overline{x}_1 - \overline{x}_2}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}}\\\\t = \dfrac{120-100}{\sqrt{20^2/22 + 30^2/15}} = \dfrac{20}{\sqrt{200/11 + 60}}\\\\\\t \approx 2.2619[/tex]

Thus, the correct test statistic for the test of the hypothesis for the considered situation is 2.2619 approximately.

Learn more about two sample t-test here:

https://brainly.com/question/23797565

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