Question

1) Use either the critical value or p-value method for testing hypotheses.
2) Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), and critical value(s).
3) State your final conclusion that addresses the original claim. Include a confidence interval as well and restate this in your original conclusion.

In a random sample of 300 patients, 21 experienced nausea. A drug manufacturer claims that fewer than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. Test this claim at the 0.05 Significance Level.

Answer 1

p: Proportion of patients who take the new drug for treating Alzheimer’s disease will experience nausea

Claim :  that fewer than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea i.e p <0.10

Null hypothesis : Ho : p = 0.10

Alternate Hypothesis : Ha: p < 0.10

Left tailed test

Hypothesized proportion : po =0.10

Number of patients in the random sample : sample size : n= 300

Number of patients of this sample who experienced nausea : x=21

Sample proportion patients who experienced nausea : = 21/300 = 0.70

Test Statistic

For left tailed test :

As P-Value i.e. is less than Level of significance i.e (P-value:0.0416 < 0.05:Level of significance); Reject Null Hypothesis

Critical value = - Z = - Z0.05 = -1.6449

As value of the test statistic: Z is less than Critical Value i.e. ( -1.7321<-1.6449 ); Reject Null Hypothesis.

There is sufficient evidence to conclude that fewer than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea

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Confidence interval

Formula for confidence interval for single population proportions : p

/2= 0.05/2 =0.025

Z/2 = Z0.025 = 1.96

95% Confidence interval for the Proportion of patients who take the new drug for treating Alzheimer’s disease will experience nausea

With 95% confidence we can conclude that 95% Confidence interval for the Proportion of patients who take the new drug for treating Alzheimer’s disease will experience nausea is (0.041127,0.098873)

As the upper confidence limit : 0.0989 < 0.10 (Hypothesized Proportion); Null hypothesis rejected;