Chapter 6 Sampling Distributions True/False 1. If we have a sample size of 100 and the estimate of the population proportion is .10, we can estimate the sampling distribution of [pic]with a normal distribution. Answer: True Difficulty: Easy 2. A sample size of 500 is sufficiently large enough to conclude that the sampling distribution of [pic] is a normal distribution, when the estimate of the population proportion is .995. Answer: False Difficulty: Medium 3.
The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2. What does the 0.2 value indicate? Answer: ____Convert 0.2 to a % by times by 100. So 0.2 = 20%, so 20% of the data is between 50 and 59______________ 3. When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6.
There is a very low probability (5% at most) that the annual income for the data from AJ DAVIS is less than $50,000. The way that we concluded this was to test the probability that the annual income of our customers is $50,000 versus the probability that the average annual income was less than $50,000. What we found was that there is a 95% chance that the average annual income of our customers is between $69,997.9 and $70,001.8. This was also backed with a p-value (which determines the strength of the evidence) that showed weak evidence against the average income equaling $50,000. Since we cannot deny that the annual income average is $50,000, we have no choice but to keep it as a consideration moving forward.
(b) One correct answer to this part is as follows: We can say that we are 95 percent confident the average balance for all overdue bills is in the range from $480 to $520. Since the standard error SE for the sample mean should equal 100/square root of 100 = 10, the margin of error would be twice this value, or $20. This means $480 to $520 is a 95 percent confidence interval for the average of all overdue balances. Another correct answer to this part is as follows: Instead of the average balance, we can find the percentage of all balances that fall in this interval. When we think of all balances, $480 is 0.2 standard deviations below the average (480 -500)/100 = - 0.20), and $520 is 0.2 standard deviations above the average (520 -500)/100 = 0.20).
Practice Problems Psy 315 Chapter 2: 11. The mean is 2, the median is 2, the sum of squared deviations is -52, the variance is -2.47, and the standard deviation is 1.57. 12. The mean is 1.361, the median is 1.3124, the sum of squared deviations is 0.07608920.0190223, the variance is 0.0190223, and the standard deviation is 0.1379213544. 13.
(d) There are two alternatives: (i) Winning $1 if the sum of the 200 numbers drawn is between –5 and +5 6 Winning $1 if the average of the 200 numbers drawn is between –0.025 and +0.025. Which is better, or are they the same? The number of draws is n=200 (a) average=sum/n = (30/200) = 0.15 (b) average=sum/n = (-20/200) = -0.1 (c) Average = sum/200 (d) (i) and (ii) are the same. –0.025 = (-5/200) and 0.025=(5/200). So if –5 happens, -0.025 is happening, too, and if 5
The alternative hypothesis H1 is that the mean annual income μ is less than $50,000. H1:μ < $50,000 Significance level chosen is 5% or α = 0.05 Here, the population standard deviation is unknown. Hence, we use a t statistic Therefore the test statistic used is t = X-μS/n follows a t distribution with n-1 degrees of freedom From the t table corresponding to 0.05 probability, the critical value tα =1.6766. Hence the critical region is t < -1.6766. Alternatively, we reject the null hypothesis, if the p value is less than the significance level Substituting the value we get t = 43.74-5014.6396/50 = -3.02 The p value corresponding to t = -3.02 and 49 d.f.
The number of terms is n=10, the first term is a1=525, the common ratio is r = 1.05. Although the initial balance is $500, a1 = $525 because the first term of the sequence is at the end of the first year, so it must include the interest on the $500. The ending balance can now be found An = a1(rn-1) A10 = 525(1.05)9 A10 = 525(1.55132822) A10 = 814.447316 814.447316 can be rounded to 814.45, thus showing that the ending balance after 10 years is $814.45. The formula to solve this problem was found on page 229 in in Mathematics in Our World (Bluman,