The average life span of Galapagos tortoises is 111 years, with a standard deviation of 29.5 years. You have done research on the average life spans of a random sample of 48 tortoises. 1) How likely is it that the average life span of a random sample of 48 tortoises is more than 120 years? A) .017 B) .355 C) .002 D) .380 2) How likely is it that the average life span of a random sample of 60 tortoises is less than 100 years? A) .017 B) .355 C) .002 D) .380 3) How likely is it that one randomly selected tortoise has a life span of more than 120 years?

The average life span of Galapagos tortoises is 111 years, with a standard deviation of 29.5 years. You have done research on the average life spans of a random sample of 48 tortoises. 1) How likely is it that the average life span of a random sample of 48 tortoises is more than 120 years? A) .017 B) .355 C) .002 D) .380 Remember that this question involves using the Central Limit Theorem. We have to adjust the standard deviation by dividing by the 𝑛 . 2nd VARS 2 (normalcdf), set your lower value to 120, your upper value to 1E99, the mean to 111, and the standard deviation to 29.5 48 . The answer is A

The average life span of Galapagos tortoises is 111 years, with a standard deviation of 29.5 years. You have done research on the average life spans of a random sample of 48 tortoises. 1) How likely is it that the average life span of a random sample of 48 tortoises is more than 120 years? A) .017 B) .355 C) .002 D) .380 NEWS FLASH!!! STAT-Test-1 works for this, too!! Use 29.5 for σ, 111 for the mean, 48 for n and designate a right tail test. The answer is your p-value, or .017 (still A).

The average life span of Galapagos tortoises is 111 years, with a standard deviation of 29.5 years. You have done research on the average life spans of a random sample of 48 tortoises. 2) How likely is it that the average life span of a random sample of 60 tortoises is less than 100 years? A) .017 B) .355 C) .002 D) .380 Again, this question involves using the Central Limit Theorem. We have to adjust the standard deviation by dividing by the 𝑛 . 2nd VARS 2 (normalcdf), set your lower value to -1E99, your upper value to 100, the mean to 111, and the standard deviation to 29.5 60 . The answer is C

The average life span of Galapagos tortoises is 111 years, with a standard deviation of 29.5 years. You have done research on the average life spans of a random sample of 48 tortoises. 2) How likely is it that the average life span of a random sample of 60 tortoises is less than 100 years? A) .017 B) .355 C) .002 D) .38 Using STAT-Test-1, use 29.5 for σ, 111 for the mean, 60 for n and designate a left tail test. The answer is your p-value, or .002 (still C).

The average life span of Galapagos tortoises is 111 years, with a standard deviation of 29.5 years. You have done research on the average life spans of a random sample of 48 tortoises. 3) How likely is it that one randomly selected tortoise has a life span of more than 120 years? A) .017 B) .355 C) .002 D) .380 Since we are only dealing with 1 tortoise, there is no need to divide by the 𝑛 2nd VARS 2 (normalcdf), set your lower value to 120, your upper value to 1E99, the mean to 111, and the standard deviation to 29.5. The answer is D. Using STAT-Test-1, use 29.5 for σ, 111 for the mean, 1 for n and designate a right tail test. The answer is your p-value, or .380 (still D).

Section 8-4 Testing the Difference Between Proportions Always assume that there is no difference between the population proportions (p1 = p2) To use a z-test to test for differences in proportions, 3 conditions must be met: 1) Samples must be randomly selected. 2) Samples must be independent. 3) np and nq must both be ≥ 5 for BOTH samples. 𝑛 1 𝑝 1 ≥5, 𝑛 1 𝑞 1 ≥5, 𝑛 2 𝑝 2 ≥5, 𝑛 2 𝑞 2 ≥5.

Section 8-4 Testing the Difference Between Proportions On the calculator, you would use STAT-TEST-6 (2-PropZTest). Use 𝑝 1 ∗ 𝑛 1 for 𝑥 1 and 𝑝 2 ∗ 𝑛 2 for 𝑥 2 (unless 𝑥 1 and 𝑥 2 are given). Make certain that your x values are whole numbers; the calculator will not accept decimals as x values. Round to the nearest whole number, if needed. Once you have the p-value, make your decision.

Section 8-4 Example 1 (Page 473) In a study of 200 randomly selected adult female and 250 randomly selected adult male Internet users, 30% of the females and 38% of the males said that they plan to shop online at least once during the next month. At α = 0.10, test the claim that there is a difference between the proportion of female and the proportion of male Internet users who plan to shop online. Write the hypotheses and identify the claim. Since the claim is that “there is a difference”, we are saying that they are not equal. H0: 𝑝 1 = 𝑝 2 Ha: 𝑝 1 ≠ 𝑝 2 (Claim) This is a two-tail test.

Section 8-4 Example 1 (Page 473) STAT – Test – 6 𝑥 1 = 𝑝 1 𝑛 1 =.3∗200=60; 𝑛 1 =200; 𝑥 2 = 𝑝 2 𝑛 2 =.38∗250=95; 𝑛 2 =250 Designate a two-tail test (≠). z = -1.775 p = .076 Make a decision (Compare p to α). Since 0.076 ≤ .10, reject the null. Since we reject the null, we support the claim. We should reject the null hypothesis and conclude that there is a difference between the proportion of female and male Internet users who plan to shop online. (At least at the 10% significance level).

Section 8-4 Example 2 (Page 474) A medical research team conducted a study to test the effect of a cholesterol- reducing medication. At the end of the study, the researchers found that of the 4700 randomly selected subjects who took the medication, 301 died of heart disease. Of the 4300 randomly selected subjects who took a placebo, 357 died of heart disease. At 𝛼=0.01, can you conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo? Write the hypotheses and identify the claim. The claim is that the proportion of heart disease deaths is lower with the medication than it is without it. H0: 𝑝 1 ≥ 𝑝 2 ; Ha: 𝑝 1 < 𝑝 2 (claim) This is a Left-tailed test

Section 8-4 Example 2 (Page 474) Run STAT Test 6 𝑥 1 =301, 𝑛 1 =4700, 𝑥 2 =357, 𝑛 2 =4300 Designate a left-tail test (<). z = -3.455 p = 2.75E-4 (.0003) Make a decision (Compare p to α). Since .0003 ≤ .01, reject the null. Since we rejected the null, we support the claim. We should reject the null hypothesis and conclude that the death rate due to heart disease is lower for those who took the medication than for those who took the placebo (at least at a 1% level of significance).

Assignments: Classwork: Pages 475-477 #3-16 Homework: Pages 477-478 #18-24 Evens Just run the test to find p and make your decision.