Question

Kate Sanders, a researcher in the department of biology at IPFW University, studied the effect of agriculture contaminants on the stream fish population in northeastern Indiana (April 2012). Specially designed traps collected samples of fish at each of four stream locations. A research question was: did the differences in agricultural contaminants found at the four locations alter the proportion of the fish population by gender? Observed frequencies were as follows.

Excel File: data12-21.xlsx

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a. Focusing on the proportion of male fish at each location, test the hypothesis that the population proportions are equal for all four locations. Use a  level of significance.

1.

2.

3.

Choose correct answer from above choice

- Select your answer -123Item 1

: - Select your answer -All population proportions are equalNot all population proportions are equalItem 2

The -value is - Select your answer -less than .005between .005 and .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 3

What is your conclusion?

- Select your answer -The population proportion of female fish equal in all four locations.The population proportion of male fish not equal in all four locations.The population proportion of female fish not equal in all four locations.The population proportion of male fish are equal in all four locations.Item 4

b. Does it appear that differences in agricultural contaminants found at the four locations altered the fish population by gender?

- Select your answer -YesNo

Answer 1

Answer:----                                                               Date:--29/3/2020

Check whether or not there are statistically significant differences among these proportions. The test hypotheses are given below: Null hypothesis: H:P = P2 = p. = P4 That is, the proportions are equal for all four locations. Alternative hypothesis: H.: Not all population proportion are equal. Thus, the correct answer is H: P. = P2 = P, = P4 Thus, the correct answer Not all population proportion are equal. Find the p-value: Compute the expected count, chi-square test and P-value by using MINITAB. The formula for Chi-square test statistic is given below: x2 = 5 (Observed – Expected) 2 2 Expected MINITAB Procedure: Step 1: Choose Stat > Tables > Chi-Square Test (Two-Way Table in Worksheet). Step 2: In Columns containing the table, enter the columns of A, B, C and D. Step 3: Click OK.

MINITAB output: Chi-Square Test: A, B, C, D Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts 1 Total 181 C D 49 44 49 39 46.81 46.81 44.21 43.17 0.102 0.169 0.519 0.403 2 167 41 46 36 44 43.19 43.19 40.79 39.83 0.111 0.183 0.563 0.436 Total 90 90 85 83 348 Chi-Sq = 2.486, DF = 3, P-value = 0.478 From the above MINITAB output, the value of Chi-square test statistic is 2.486 and the P-value is 0.478 The p-value 0.478 is greater than the 0.10. Thus, the p-value lies between the greater than 0.10 Conclusion: Use significance level, a=0.05. Here, P-value is 0.478 which is greater than the value of a = 0.05. That is. (P-value = 0.478)>(a=0.05). Therefore, by the rejection rule, it can be concluded that there is evidence to do not reject H, at a=0.05. Hence, the population proportion of male fish are equal in all four locations.

In this test the null hypothesis indicates that the population proportions of male fish at four locations are all equal. Thus, there is Does not appear that differences in agricultural contaminants found at the four locations altered the fish population by gender. Thus, the correct answer is No. Thus, the No does not appear that differences in agricultural contaminants found at the four locations altered the fish population by gender.