A cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. You work for a national health service and are asked to test this claim. You find that a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg. For a significance level of 0.05, do you have enough evidence to reject the company’s claim?

a. Find the value of the test statistic
b. Find the p-value

We are given that a cereal company claims the mean sodium content in one serving of its cereal is no more than 230 mg. For this a random sample of 52 servings has a mean sodium content of 232 mg and a standard deviation of 10 mg.

We have to test whether company's claim was true or not.

Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] 230 mg {means that the company claims of mean sodium content in one serving of its cereal is no more than 230 mg is true}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 230 mg {means that the company claims the mean sodium content in one serving of its cereal is no more than 230 mg is not true}

(a) The test statistics that will be used here is One sample t-test statistics;

T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, Xbar = sample mean sodium content = 232 mg

s = sample standard deviation = 10 mg

n = sample of servings = 52

So, test statistics = [tex]\frac{232-230}{\frac{10}{\sqrt{52} } }[/tex] ~ [tex]t_5_1[/tex]

= 1.442

Therefore, the value of test statistics is 1.442 .

(b) Now, the P-value is given by = P([tex]t_n_-_1[/tex] > test statistics)

= P([tex]t_5_1[/tex] > 1.442) = 0.081 or 8.1%

Since, P-value is greater than the significance level as 8.1% > 5%, so we have insufficient evidence to reject null hypothesis and conclude that company's claim was true.