Researchers want to know about the true proportion of adults with at least a high schooleducation. 1000 adults are surveyed, and 710 of them have at least a high school education. create a 95% confidence interval for the true population proportion of adults with at least a highschool education.

(a) H₀: μ ≥ 5.70 vs. Hₐ:μ

(b) The rejection region is (t₀.₀₁,₉ ≤ -2.821).

(c) The value of the test statistic is -4.33.

(d) The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

Step-by-step explanation:

A hypothesis test should be conducted to determine that the if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(a)

The hypothesis is:

H₀: The true mean breaking strength of the new bonding adhesive is not less than 5.70 Mpa, i.e. μ ≥ 5.70.

Hₐ: The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, i.e. μ

(b)

The alternate hypothesis indicates that the hypothesis test is left-tailed.

The rejection region for the left tailed test will be towards the lower tail of the t-distribution curve.

The significance level of the test is: α = 0.01.

The critical value is:

Use the t-table for the critical value.

Since rejection region is in the lower tail the critical value will be negative.

Thus, the rejection region is (t₀.₀₁,₉ ≤ -2.821).

(c)

The test statistic value is:

Given:

Compute the value of the t-statistic as follows:

The value of the test statistic is -4.33.

(d)

The value of the test is less than the critical value.

This implies that the test statistic lies in the rejection region.

Hence the null hypothesis will be rejected at 1% significance level.

Conclusion:

As the null hypothesis is rejected it can be concluded that the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(e)

The conditions required for the t-test for single mean to be valid is:

The data should be continuous.The parent population should be normally distributed.The sample should be randomly selected.

   

 

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is less than 5.7 Mpa at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

represent the sample mean

represent the sample standard deviation

sample size  

represent the value that we want to test

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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 true mean is less than 5.7 Mpa, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

If we analyze the size for the sample is

 (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:  

   

P-value

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

 

Since is a one side lower tailed test the p value would be:  

 

Conclusion  

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is less than 5.7 Mpa at 1% of signficance.  

   

 

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the treu mean is different from 30 minutes at 1% of significance

Step-by-step explanation:

Data given and notation  

represent the sample mean

represent the sample standard deviation

sample size  

represent the value that we want to test

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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

1) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 30 minutes, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

If we analyze the size for the sample is > 30 but 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:  

 (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".  

2) Calculate the statistic

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

   

3) P-value

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

 

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

 

4) Conclusion  

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the treu mean is different from 30 minutes at 1% of significance

The 95% confidence interval for the true population proportion of adults with at least a high school education is (0.6819, 0.7381). This means that we are 95% sure that the true proportion of adults in the entire population surveyed with at least a high school education is (0.6819, 0.7381).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

95% confidence level

So , z is the value of Z that has a pvalue of , so .  

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the true population proportion of adults with at least a high school education is (0.6819, 0.7381). This means that we are 95% sure that the true proportion of adults in the entire population surveyed with at least a high school education is (0.6819, 0.7381).

I think it’s true but I could be wrong