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Using specific data, we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. This exercise asks you to conduct a hypothesis test using additional data. In this case, we are testing

??????H0??:?po?=?pcHa?:?po??>?pc

where po and pc represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Use a 5% significance level.

Effect of Organic Potatoes After 11 Days

After 11 days, the proportion of fruit flies eating organic potatoes still alive is 0.67, while the proportion still alive eating conventional potatoes is 0.63. The standard error for the difference in proportions is 0.033.

What is the value of the test statistic?

Round your answer to two decimal places.

?z=?

What is the p-value?

Round your answer to three decimal places.

p-value=

What is the conclusion?
Reject H_0.Do not reject H_0.



Is there evidence of a difference?

Respuesta :

akindeleot

Answer:

Test statistic:

z = (0.67 - 0.63) / 0.033 = 1.21

p - Value = P(z > 1.21) = 0.106

My conclusion: Fail to reject null hypothesis

No, there is no evidence of a difference

The value of the test statistic is 1.38The p-value is 0.0838.

So, there is no evidence of difference.

To understand the calculations, check below

Test Statistic:

In a test of hypothesis, the test statistic is a function of the sample data used to decide whether or not to reject the null hypothesis.

Given that, the null and alternative hypotheses are,

[tex]H_0:p_0=p_c\\H_a:p_0 > p_c[/tex]

This hypothesis test is a right-tailed test.

The proportion of fruit flies eating organic potatoes still alive is 0.67, while the proportion still alive eating conventional potatoes is 0.63. The standard error of the difference in proportions is 0.029.

The value of the test statistic is,z

[tex]z=\frac{\hat{p_1}-\hat{p_2}}{SD_{\hat{p_1}-\hat{p_2}}}[/tex]

Now, substituting the given values into the above formula we get,

[tex]\frac{0.67-0.63}{0.029} =1.38\\z=1.38[/tex]

Now, calculating the p-value as,

[tex]p-value = P(Z > 1.38) \\= 1 - P(Z < 1.38)\\ = 1 - 0.9162\\ = 0.0838[/tex]

Since the p-value is greater than the 0.05 significance level, we fail to reject the null hypothesis [tex]H_0[/tex].

Learn more about the topic Test Statistic:

brainly.com/question/25629572

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