Mathematics, 07.03.2020 03:52 Chandler1Gaming

The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs. He decides that he wants to be fairly sure that the mean weight of these ice cream tubs (μ) is greater than 52 ounces. A hypothesis test is conducted with the following hypotheses: H0: μ = 52 Ha: μ > 52 The level of significance used in this test is α = 0.05. A random sample of 26 ice cream tubs are collected and weighed. The sample mean weight is calculated to be x = 54.98 and the sample standard deviation is s = 8.43. a)Calculate the test statistic (t) for this test. Give your answer to 4 decimal places.

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Answer from: pnhandley01

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

We need to find the degrees of freedom given by:

df = n-1= 26-1 = 25

Since is a right tailed test the p value would be:

$p_v =P(t_{25}1.8025)=0.0418$

If we compare the p value and the significance level given $\alpha=0.05$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=54.98$ represent the sample mean

s=8.43 represent the sample deviation

n=26 sample size

$\mu_o =52$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:

Null hypothesis: $\mu \leq 52$

Alternative hypothesis: $\mu 52$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025$

P-value

We need to find the degrees of freedom given by:

df = n-1= 26-1 = 25

Since is a right tailed test the p value would be:

$p_v =P(t_{25}1.8025)=0.0418$

Conclusion

Answer from: Quest

the answer is 8

step-by-step explanation:

3.14 x 8x8

Answer from: Quest

where are the options?

step-by-step explanation:

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