Arandom sample of 65 bags of white cheddar popcorn weighed, on average, 5.74 ounces with a standard deviation of 0.26 ounce. test the hypothesis that muequals5.8 ounces against the alternative hypothesis, muless than5.8 ounces, at the 0.10 level of significance.

We failed to accept null hypothesis

Step-by-step explanation:

x= 5.74

Standard deviation =

n = 65

We will use z test

Formula :

Substitute the values :

Refer the z table for p value

p value = 0.0314

α = 0.10

p value < α

So, we failed to accept null hypothesis

step-by-step explanation:

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;