The average time customers spent on the American Greetings Web site has been 11.85 minutes (Top Web Properties, USA Today, April 27, 2000). You want to show that the average time spent on that web site has now changed. Assume that the population of site visit times is normally distributed with a population standard deviation of 4.0 minutes. In order to test the hypothesis at 0.05 level of significance, the critical value(s) is(are):

± 1.96

Step-by-step explanation:

Given the following :

Mean (m) time spent = 11.85

Standard deviation = 4 minutes

Since the population of site visits is Normally distributed, Hence, the test statistic will follow a normal distribution :

The test the hypothesis that mean time has changed at 0.05 confidence interval

Normal distributions have a mean value of 0 and standard deviation of 1

Using a two tailed test At 0.05 confidence interval ;

Using the critical value calculator, which in general under normal distributions :

0.05% confidence interval for a two tailed test is

± 1.96

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

answer: are you

step-by-step explanation: