Significance Test for a Mean Section 9.2 Day 2 Significance Test for a Mean

Dealing with Outliers Sometimes your data will contain outliers. How should you deal with these?

Dealing with Outliers So, do the analysis twice There might be a good reason to remove the outliers. So, do the analysis twice

Dealing with Outliers So, do the analysis twice once with the outliers There might be a good reason to remove the outliers. So, do the analysis twice once with the outliers

Dealing with Outliers So, do the analysis twice once with the outliers There might be a good reason to remove the outliers. So, do the analysis twice once with the outliers once without them

Dealing with Outliers If you are fortunate . . .

Dealing with Outliers If you are fortunate, your analysis will lead you to the same conclusion both times . . . either reject or do not reject the null hypothesis.

Dealing with Outliers If you have a “split decision”, then . . .

Dealing with Outliers If you have a “split decision”, then you’ll need more data.

Dealing with Outliers If you have a “split decision”, then you’ll need more data. You do not want your conclusion to depend on what you assume about one or two observations.

Page 597, P24 Name the test

Name of Test One-sided significance test of a mean One-sided because “you are convinced that the room tends to be colder.”

Check Conditions

Check Conditions 1) Problem states the temperatures were taken on randomly selected days.

Check Conditions 2) The plot of the data is fairly symmetrical with no outliers, so it’s reasonable to assume the sample came from a normal population. (You must show a plot.)

Check Conditions 3) The population of days on which the temperatures could be taken is more than 10 times the sample size, (10x7 = 70). So, all conditions met.

Write Hypotheses

Write Hypotheses Ho: μ = 72o F, where μ is the mean temperature at my seat

Write Hypotheses Ho: μ = 72o F, where μ is the mean temperature at my seat Ha: μ < 72o F (because “you are convinced that the room tends to be colder.”)

Computations

Computations T-Test < 72 t = - 2.931 p = .0131

Conclusion I reject the null hypothesis because the P-value of 0.0131 is less than the significance level of α = 0.05.

Conclusion I reject the null hypothesis because the P-value of 0.0131 is less than the significance level of α = 0.05. There is sufficient evidence to support the claim that the mean temperature at your desk is less than 72o F.

Page 587, D9

Page 587, D9 a) If s increases while everything else remains the same, the absolute value of the test statistic, t, will decrease. Thus, the P-value will ________.

Page 587, D9 a) If s increases while everything else remains the same, the absolute value of the test statistic, t, will decrease. Thus, the P-value will increase.

Page 587, D9

Page 587, D9 b) If the sample size increases but everything else remains the same, the absolute value of the test statistic, t, will increase. Thus, the P-value will __________.

Page 587, D9 b) If the sample size increases but everything else remains the same, the absolute value of the test statistic, t, will increase. Thus, the P-value will decrease.

Page 596, P12

Page 596, P12 The red curve is the standard normal distribution. The standard normal distribution has less area in the tails. The t-distribution is more spread out with heavier tails.

Page 599, E23

Page 599, E23 One-sided significance test for the mean because the consumer group says the average weight is less than 10 oz.

Page 599, E23 Randomness: problem states bags of chips randomly selected

Page 599, E23 Randomness: bags of chips randomly selected Normality: company claims that the weight of all bags is normally distributed

Page 599, E23 Randomness: bags of chips randomly selected Normality: company claims that the weight of all bags is normally distributed Boxplot of data shows . . . .

Page 599, E23 Boxplot of data shows distribution is fairly symmetric with no outliers so reasonable to assume sample came from normally distributed population

Page 599, E23 All conditions met. Population size: Population of bags of chips produced would be more than 150 bags which is 10 times the sample size. All conditions met.

Page 599, E23 Ho: μ = 10 oz where μ is the mean weight of bags of Munchie’s Potato Chips

Page 599, E23 Ho: μ = 10 oz where μ is the mean weight of bags of Munchie’s Potato Chips Ha: μ < 10

Page 599, E23 TTest Inpt: Data μo: 10 List: L1 Freq: 1 μ: < μo Calculate

Page 599, E23 t = - 2.55 P-value = .012

Page 599, E23 I reject the null hypothesis because the P-value of 0.012 is less than the significance level of α = 0.05.

Page 599, E23 I reject the null hypothesis because the P-value of 0.012 is less than the significance level of α = 0.05. There is sufficient evidence to support the claim that the mean weight of bags of Munchie’s Potato Chips is less than 10 oz.

Page 597, P18

Page 597, P18 a) Ho: = 1700, where is the mean SAT score of State University students

Page 597, P18 a) Ho: = 1700, where is the mean SAT score of State University students Ha:

Page 597, P18 b) TTest Inpt: Stats : 1700 x : 1535 sx: 250 n: 9

Page 597, P18 b) t = ± 1.98, P-value = .083 10% level:

Page 597, P18 b) t = ± 1.98, p-value = .083 10% level: reject because .083 < .10 5% level:

Page 597, P18 b) t = ± 1.98, p-value = .083 10% level: reject because .083 < .10 5% level: do not reject because .083 > .05 1% level:

Page 597, P18 b) t = ± 1.98, p-value = .083 10% level: reject because .083 < .10 5% level: do not reject because .083 > .05 1% level: do not reject because .083 > .01

Questions?