18 of40 (33 A genetic expertment involving peas yiekded one sample of oflspring

Question

18 of40 (33 A genetic expertment involving peas yiekded one sample of oflspring consisting of 441 151 yellow peas. Use a 0 05 signifticance level to test the claim that under the same circumstanoes 24% of oitspring peas will be yellow identify the null h othesis, alternative hypothests test statistic P value, conclusion about the null hypothesis, and final conckusion that addresses the original claim tion as an approximation to the binomial distribution mal distribu What are the null and alternative hypotheses? OA, Ho : p#024 ??. ??.p#024 OE. Ho p-024 H4 p 0.24 Hi p 024 OB, Ho: p:024 H1 ps024 OD, Hop#024 Ht p 024 1: p#024 What is the test statistic? (Round to two decimai places as needed ) Click to select your answers) 1048 PM ^???400 6/27/2018

Explanation / Answer

The statistical software output for this problem is:

One sample proportion summary hypothesis test:
p : Proportion of successes
H0 : p = 0.24
HA : p ? 0.24

Hypothesis test results:

Hence,

Hypotheses: Option F is correct.

z = 0.86

p - Value = 0.3907

Conclusion: Option D is correct.

Final conclusion: Option C is correct.

Proportion Count Total Sample Prop. Std. Err. Z-Stat P-value p 151 592 0.25506757 0.017553009 0.85840371 0.3907

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