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(Solved): A 95% confidence intervals for the proportion of first-year statistics students with brown hair is ...




A \( 95 \% \) confidence intervals for the proportion of first-year statistics students with brown hair is given to be \( (0.
A company creates wooden furniture to be sold at various retailers. This company claims that only \( 2.5 \% \) of their furni
pieces are deiective. Assume anumuar sampung
distribution.
(a) What are the null and alternative hypotheses?
\( \mathrm{H}_{0
A confidence intervals for the proportion of first-year statistics students with brown hair is given to be . (a) What is the sample proportion of students with brown hair? (Round to 3 decimal places, if needed.) (b) What is the margin of error of this interval? (Round to 3 decimal places, if needed.) (c) What is the critical value used for this interval? (Round to 3 decimal places, If needed.) (d) What is the standard error of the estimate? (Round to 3 decimal places, if needed.) (e) What was the sample size? (Round to the appropriate integer.) A company creates wooden furniture to be sold at various retailers. This company claims that only of their furniture are defective in some manner. Monica would like to test this claim to see if the actual percentage of furniture that is defective is different than the company claims, using . She takes a sample of 815 pieces of furniture, and observes that 38 of these furniture pieces are defective. Assume a normal sampling distribution. pieces are deiective. Assume anumuar sampung distribution. (a) What are the null and alternative hypotheses? : : (b) What is the test statistic? (Round your answer to 2 decimal places, if needed.) (c) Using the statistical table, what is the -value? (Round your answer to 4 decimal places, if needed.) (d) Based on the -value, Monica should the null hypothesis. (e) This data sufficient evidence to conclude that the actual percentage of furniture that is defective is different than the company claims.


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Here from the given details,The 95% confidence interval is,95%CI=(0.27,0.44)
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