Tables and Formulas To calculate test statistic z Z x o n x z n To find confidence interval when sigma is given To find the right sample size n z m To find alpha To find margin of error z n 1 C c is the confidence level in decimal form Tips for Testing We are allowed to use one notecard so try and fit as many example problems on the notecard as possible I do one example for each formula It is important to have formulas but they are not useful if you don t understand when to use them Chapter 6 Example 1 For questions 1 3 use the following information You want to rent an unfurnished one bedroom apartment for next semester The mean monthly rent for a random sample of 50 apartments is 640 Assume that the population of all unfurnished one bedroom apartments follows a Normal distribution with standard deviation is 90 The first thing to do is evaluate which given information is relevant So x bar equals 640 sigma is 90 the sample number is 50 1 What is the 95 confidence interval for the mean monthly rent of all unfurnished one bedroom apartments In order to solve this problem we need the formula z n What does this mean it means that the 95 confidence interval is equal to x bar plus or minus the z statistic multiplied by sigma divided by the square root of the random sample To find z refer to table D look at the bottom and find 95 because it is the confidence level and the number above is the z Now just plug and chug z n 640 1 96 90 50 640 24 95 The confidence interval is 615 05 664 95 2 What is the 99 confidence interval for This problem is asking the same question as the previous so we need the formula z n Note that there is a difference confidence level To find z refer to table D look at the bottom and find 99 because it is the confidence level and the number above is the z Now just plug and chug z n 640 2 576 90 50 640 32 79 The confidence interval is 607 21 672 78 3 Suppose you want a 90 confidence interval where the margin of error is m 50 What sample size do you need to construct this confidence interval m which means the sample size is The formula used for this problem is n z equal to z statistic multiplied by sigma and divided by the margin of error squared Now just plug and chug m n z 1 65 90 50 N 8 77 n 9 is the final answer Because a person cannot visit 8 77 apartments you must round to the next whole number Whenever you are finding n you must round to the higher whole number so you don t under sample 6 2 Examples 1 The Survey of Study Habits and Attitudes SSHA is a psychological test that measures the motivation attitude toward school and study habits of students Scores range from 0 to 200 The scores for all U S college students are normally distributed with a mean of 115 and a standard deviation of 30 a A teacher suspects that older students 30 or older have better attitudes toward school and thinks that they will score higher on the SSHA test than all U S college students State the appropriate null and alternative hypotheses for a significance test of the mean SSHA score of all older students The null hypothesis is the mean equals 115 so H 0 115 The alternative hypothesis is the mean is greater than fifteen so H a 115 This means it is a one sided test b The teacher gives the SSHA test to a random sample of 25 older students and their mean score is 133 2 Assuming that 30 for the population of older students and the scores are normally distributed calculate the appropriate test statistic for the hypotheses in part a For this problem the formula is Z x o n Z 133 2 115 30 25 Z 3 03 b Draw a sketch illustrating the P value and report the P value of the test c Is the result significant at the 1 level 0 01 Because the p value 0012 is less than alpha 01 we reject the null hypothesis This means the result is significant at the 1 level 2 You are interested in testing whether or not vigorous exercise affects concentration levels In general the time needed for people to complete a certain paper and pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 3 seconds You wish to see if the mean time after vigorous exercise is different from the regular time a State the appropriate null and alternative hypotheses for a significance test of the mean maze time after vigorous exercise H0 30 Ha 30 this is a two sided test b You select a random sample of 36 people and have them exercise vigorously for 30 minutes before completing the maze It takes them an average of 31 2 seconds to complete the maze Assuming that the times follow a normal distribution and the standard deviation is 3 calculate the appropriate test statistic for the hypotheses in part a Z x o n Z 31 2 30 3 36 Z 2 4 c Draw a sketch illustrating the P value and report the P value of the test d Is the result significant at the 1 level 0 01 What about at the 5 level 0 05 When alpha is at the 1 level we do not reject because the p value 0164 is greater than alpha This means that the mean time is not significantly different from 30 seconds at this level When alpha is 05 we do reject the null hypothesis because the p value 0164 is less than 05 This means that the mean time is significantly different from 30 seconds at this level 3 A 95 confidence interval for a population mean is 16 32 a Can you reject the null hypothesis that 34 at the 5 significance level in a two sided test You can reject the null hypothesis because 34 is not in the range of 16 and 32 b Can you reject the null hypothesis that 17 at the 5 significance level in a two sided test You do not reject because it is in the range of 16 and 32 Homework Problems 6 5 Health insurance For the 453 large firms in the Kaiser Health coverage survey Example 6 5 the mean monthly cost of the premium for single coverage HMO plans per worker was 405 02 The sample standard deviation of premium costs was 112 08 Assume that the sample standard deviation can be used in place of the population standard deviation We have already computed an approximate 95 confidence interval for the mean cost Construct more precisely the 95 confidence interval for the mean premium cost for all large firms Given n …