A manufacturer of metal fasteners expects to ship an average of 1197.00 boxes of fasteners per day. A random sample of 24 days produced a mean of 1193.69 boxes per day with a standard deviation of 3.3166 boxes per day. Does evidence exist showing the average number of boxes shipped per day is different from 1197.00, using alpha=5%? If it is different from 1197.00 then the manufacturer stop

The p-value obtained is less than the significance level at which the test was performed, hence, we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Step-by-step explanation:

For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that there is no evidence showing that the average number of boxes shipped per day is different from 1197.00. That is, there is no significant difference between the number of boxes shipped per day and 1197.00

And the alternative hypothesis is that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Mathematically, the null hypothesis is

H₀: μ₀ = 1197.00

The alternative hypothesis is

Hₐ: μ₀ ≠ 1197.00

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 1193.69 boxes per day

μ₀ = standard that we're comparing the sample mean with = 1197.00

σₓ = standard error of the sample mean

= (σ/√n)

where n = Sample size = 24 days

σ = standard deviation of the sample = 3.3166 boxes per day

σₓ = (3.3166/√24) = 0.677

t = (1193.69 - 1197) ÷ 0.677

t = -4.89

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 24 - 1 = 23

Significance level = 0.05

p-value (for t = -4.89, at 0.05 significance level, with a two tailed condition) = 0.000061

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.000061

0.000061 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Hope this Helps!!!

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.

do your own work and don"t cheat.

step-by-step explanation: