Sarah claims that the thickness of the spearmint gum she produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of 10 pieces of gum and measured their thickness. He obtained: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5 and 7.5. Which of the following statements is true:a. State the null and alternate hypothesis using statistical notation?H0: μ = 7.5 HA: μ ≠ 7.5b. Is this a one-tailed or a two-tailed test?Two tailed test because the specialist simply tests for difference in the claim

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis.

Step-by-step explanation:

Data given and notation

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And the sample deviation given by:

[tex] s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

And we got:

[tex]\bar X=7.55[/tex] represent the sample mean

[tex]s=0.103[/tex] represent the sample standard deviation

[tex]n=10[/tex] sample size

[tex]\mu_o =7.5[/tex] represent the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 equal to 7.5 or no, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 7.5[/tex]

Alternative hypothesis:[tex]\mu \neq 7.5[/tex]

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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:

[tex]t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539[/tex]

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n-1=10-1=9[/tex]

Since is a two sided test the p value would be:

[tex]p_v =2*P(t_{(9)}>1.539)=0.158[/tex]

Conclusion

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis.