Time Remaining: 00 54 23 This Test: Questi of offspring consisting of 446 green

Question

Time Remaining: 00 54 23 This Test: Questi of offspring consisting of 446 green peas and 120 yellow peas. Use a 0.05 significance level to test the claim that u est: Test Chapter 8 4of9(5 complete) is Question: 1 pt one s genetic rai st es, 23% of oms n g peas will be yellow Identity the null hypothesi alternative hypothess, test statistic P-value, condusion about the nu hat addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution hypo ess and final O B. Ho: p40 23 H1 :p=0.23 OD. Ho p-0.23 H1 p 023 OF. H p 0.23 H1 : pt023 H p>023 Ho p#023 H, p023 OE Ho p -0.23 H Px023 What is the test statistic? Round to two decimal places as needed.) What is the P-value? P-value Round to four decimal places as needed) What is the conclusion about the null hypothesis? A. Fat to reject the null hypothesis because the P-value is greater than the significance level, B. Reed the null hypothesis because the P-value is greater than the significance level Click to select your answerls)

Explanation / Answer

as the claim is that 23% will be yellow
hence this alternate hypothesisis
p=0.23 and null hypothesis is p != 0.23 , hence B

446 greena and 120 yellow
total = 446+120
= 566

so the proportion of yellow is
120/566 = 0.212

now we know that the Z stat is given as

z= (phat - p)/(sqrt(p*1-p)/n)
= (0.212-0.23)/(sqrt(0.23*0.77/566)) = -1.017

so we check the z table to find the corresponding p value as 0.1545

as the p value is greater than 0.05 , hence we fail to reject the null hypothesis

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