A marketing research firm suspects that a particular product has higher name recognition among college graduates than among high school graduates. A sample from each population is selected, and each asked if they have heard of the product in question. A summary of the sample sizes and number of each group answering “yes” are given below:

College Grads (Pop. 1) : n1 = 93, x1 = 61
High School Grads (Pop. 2) : n2 = 98, x2 = 58

Is there evidence, at a α = 0.013 level of significance, to support the claim of higher name recognition among college graduates? Carry out an appropriate hypothesis test, filling in the information requested.
A. The value of the standardized test statistic:

B. The p-value is

C. Your decision for the hypothesis test:

*******How do I do this without standard devations? I was going to do a hypothesis test with two independent means but how can I?

*****ALSO**** what is the formula to find p values? I usually just compare them to the Z score andt eh Z star...

To perform the hypothesis test without standard deviations, you can use a two-sample z-test for proportions. Here's how you can proceed with the given information:

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The name recognition proportions for college graduates and high school graduates are equal.
- Alternative hypothesis (Ha): The name recognition proportion for college graduates is higher than that for high school graduates.

Step 2: Calculate the sample proportions:
- Sample proportion for college graduates: p1 = x1/n1 = 61/93 ≈ 0.656
- Sample proportion for high school graduates: p2 = x2/n2 = 58/98 ≈ 0.592

Step 3: Calculate the standard error:
The formula for the standard error of the difference in proportions is:
SE = sqrt( (p1*(1 - p1))/n1 + (p2*(1 - p2))/n2 )

SE = sqrt( (0.656*(1 - 0.656))/93 + (0.592*(1 - 0.592))/98 )

Step 4: Calculate the test statistic:
The test statistic, which approximates the standard normal distribution, is given by:
z = (p1 - p2) / SE

z = (0.656 - 0.592) / SE (Substitute the values of p1, p2, and SE)

Step 5: Calculate the p-value:
To calculate the p-value, you need to compare the absolute value of the test statistic to the standard normal distribution. The p-value is the probability of obtaining a test statistic more extreme than the one observed.

You can use statistical software or tables to find the p-value corresponding to the calculated test statistic (z-value). In this case, the p-value is the probability of obtaining a z-value greater than the calculated test statistic.

Step 6: Make a decision based on the p-value and significance level:
- If the p-value is less than the chosen significance level (α = 0.013), we reject the null hypothesis in favor of the alternative hypothesis. There would be evidence to support the claim of higher name recognition among college graduates.
- If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis. There would not be sufficient evidence to support the claim of higher name recognition among college graduates.

Regarding your second question, the formula to find p-values depends on the statistical test being used. For standard normal distribution (z-test), you can calculate the p-value by finding the area under the curve using tables or statistical software. p-values can also be calculated for other tests like t-tests or chi-square tests using their respective distributions.

Note: It's important to keep in mind that this explanation assumes certain conditions are met for the z-test to be appropriate, such as random sampling and independence of the samples.