How many tissues should the kimberly clark corporation package of kleenex contain? researchers determined that 60 tissues is the mean number of tissues used during a cold. suppose a random sample of 100 kleenex users yielded a mean of 52 tissues used during a cold with a standard deviation of 22. suppose the alternative you wanted to test was h1: µ < 60. if the critical value in this situation at an α = .05 is -1.6604, when should we reject the null?

Step-by-step explanation:

For the null hypothesis,

µ = 60

For the alternative hypothesis,

h1: µ < 60

This is a left tailed test

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100,

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 52

µ = population mean = 60

s = samples standard deviation = 22

t = (52 - 60)/(22/√100) = - 3.64

We would determine the p value using the t test calculator. It becomes

p = 0.00023

We would reject the null hypothesis if α = 0.05 > 0.00023

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;