Hippocrates magazine states that 32 percent of all Americans take multiple vitamins regularly. Suppose a researcher surveyed 750 people to test this claim and found that 261 did regularly take a multiple vitamin. Is this sufficient evidence to conclude that the actual percentage is different from 32% at the 5% significance level? Select the [p-value, Decision to Reject (RHo) or Failure to Reject (FRHo)1.
a) [p-value = 0.069, FRHI
b) [p-value = 0.009, RH01
c) [p-value = 0.009, FRHol
d) [p-value = 0.019, FRH)]
e) [p-value = 0.019, RHo]

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

p = 0.32

For the alternative hypothesis,

p ≠ 0.32

This is a two tailed test

Considering the population proportion, probability of success, p = 0.32

q = probability of failure = 1 - p

q = 1 - 0.32 = 0.68

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 261

n = number of samples = 750

P = 261/750 = 0.35

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.35 - 0.32)/√(0.32 × 0.68)/750 = 1.8

Recall, population proportion, p = 0.32

The difference between sample proportion and population proportion(P - p) is 0.35 - 0.32 = 0.03

Since the curve is symmetrical and it is a two tailed test, the p for the left tail is 0.32 - 0.03 = 0.29

the p for the right tail is 0.32 + 0.03 = 0.35

These proportions are lower and higher than the null proportion. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area above the z score in the right tail 1 - 0.9641 = 0.0359

We would double this area to include the area in the right tail of z = 0.44 Thus

p = 0.0359 × 2 = 0.07

Since alpha, 0.05 < the p value, 0.07 then we would fail to reject the null hypothesis. Therefore, this is not sufficient evidence to conclude that the actual percentage is different from 32% at the 5% significance level.

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;