1997, a survey of 940 house holds showed that 15 of them use e-mail. Use those sample results to test the claim that more than 15 % of households use email. use a 0.05 significance level.

which of the following is the hypothesis test to be conducted

a. h0 :p=0.15
h1:p> 0.15

b h0: p<0.15
h1; p=0.15

c. ho>0.15
h1 =0.15

d. h0: p/= 0.15
h1 p=0.15

e h0 p=0.15
h1 p<0.15

f. H0:p =0.15
h1:p?=0.15

what is the test statistic? z=
what is the p-value
d. what is the conclusion?
a. There is not sufficient evidence to support the claim that more than 15% of households use e-mail.
b there is sufficient evidence to support the claim that more than 15%of households use email

e. is the conclusion valid today? why or why not?
a. yes the conclusion is valid today because the requirements to perform the test are satisfied.
b. no the conclusion is not valid today because the population characteristics of the use of email are changing rapidly.
c. you can make no decisions about the validity of the conclusion today.

The hypothesis test to be conducted is:

a. H0: p = 0.15
H1: p > 0.15

The test statistic is the z-score. We can calculate the test statistic using the formula:

z = (p̂ - p) / √(p(1 - p) / n)

where:
p̂ = proportion of households in the sample using email (15/940 = 0.016)
p = proportion claimed in the hypothesis (0.15)
n = sample size (940)

Plugging in the values, we get:

z = (0.016 - 0.15) / √(0.15 * (1 - 0.15) / 940)

The p-value is the probability of observing a test statistic as extreme as the one calculated assuming the null hypothesis is true. We can find the p-value by looking up the z-score in the standard normal distribution table.

If the p-value is less than the significance level (0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Once the p-value is obtained, we can draw a conclusion.

d. The conclusion is:
a. There is not sufficient evidence to support the claim that more than 15% of households use e-mail.

e. The conclusion may not be valid today because the conclusion is based on the survey results from 1997. The population characteristics of the use of email could have changed since then, potentially invalidating the conclusion.

The correct hypothesis test to be conducted is:

a. H0: p = 0.15
H1: p > 0.15

Here, H0 represents the null hypothesis, which states that the proportion of households using e-mail is equal to 15%. H1 represents the alternative hypothesis, which states that the proportion is greater than 15%.

The test statistic in this case is the z-score. To calculate the z-score, you need to use the sample proportion (15/940) and the population proportion under the null hypothesis (0.15). The formula for the z-score is:

z = (p̂ - p0) / sqrt(p0 * (1 - p0) / n)

where p̂ is the sample proportion, p0 is the population proportion under the null hypothesis, and n is the sample size.

To find the p-value, you need to compare the z-score to the standard normal distribution. Since the alternative hypothesis is one-tailed (p > 0.15), you need to find the area under the curve to the right of the z-score. This can be done using a z-table or a statistical software.

To interpret the results, compare the p-value to the significance level (0.05). If the p-value is less than the significance level, you reject the null hypothesis. If the p-value is greater than or equal to the significance level, you fail to reject the null hypothesis.

In this case, if the p-value is less than 0.05, it would indicate sufficient evidence to support the claim that more than 15% of households use email. Therefore, the correct conclusion would be:

b. There is sufficient evidence to support the claim that more than 15% of households use email.

To determine if the conclusion is valid today, you need to consider if the population characteristics of the use of email have significantly changed since the survey was conducted in 1997. This information is not provided in the question. Therefore, the correct answer would be:

c. You can make no decisions about the validity of the conclusion today.

Use a) for your hypotheses.

You can try a proportional one-sample z-test since this problem is using proportions.

Using a formula for a proportional one-sample z-test with your data included, we have:
z = .16 - .15 -->test value (150/940) is .16) minus population value (.15) divided by
√[(.15)(.85)/940] --> .85 represents 1-.15 and 940 is sample size.
Finish the calculation.
Use a z-table to find the p-value. The p-value is the actual level of the test statistic.

Check a z-table for the critical value at .05 level of significance for a one-tailed test. Compare the test statistic you calculated to the critical value from the table. If the test statistic exceeds the critical value, reject the null and conclude p>0.15 (there is sufficient evidence to support the claim); if the test statistic does not exceed the critical value from the table, do not reject the null (there is not sufficient evidence to support the claim).

I'll let you take it from here.

In​ 1997, a survey of 860


households showed that 144

of them use​ e-mail. Use those sample results to test the claim that more than​ 15% of households use​ e-mail. Use a 0.05 significance level. Use this information to answer the following questions.