In a clinical trial, out of patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that % of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than % of this drug's users experience flulike symptoms as a side effect at the level of significance?

Step-by-step explanation:

Hello!

Out of 846 patients taking a prescription drug daily, 18 complained of flulike symptoms.

It is known that the population proportion of patients that take the drug of the competition and complain of flulike is 1.8%

Be the variable of interest:

X: number of patients that complained of flulike symptoms after taking the prescription drug, out of 846.

sample proportion p'= 18/846= 0.02

You have to test if the population proportion of patients that experienced flulike symptoms as a side effect is greater than 1.8% (p>0.018)

Assuming that the patients for the clinical trial were randomly selected.

The expected value for this  sample is np=846*0.02= 1658 (the expected value of successes is greater than 10) and the sample is less than 10% of the population, you can apply the test for the proportion:

The hypotheses are:

H₀: p ≤ 0.018

H₁: p > 0.018

α: 0.01

≈N(0;1)

The p-value for this test is  0.331056

The decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value is greater than α, the decision is to reject the null hypothesis.

So at 1% significance level there is no significant evidence to reject the null hypothesis, you can conclude that the population proportion of patients that took the prescription drug daily and experienced flulike symptoms as a side effect is less or equal to 1.8%

I hope this helps!

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;