# Statistics and Probability Practice Problems

1)Fill in the blanks or circle the correct answer for the following: a)If P(A∩B)=P(A)P(B) then A and B are called **independent.** b)Write the type of the variables i) Student’s N# **qualitative** ii) Student’s age **quantitative-discrete** c)If we reject Ho at 5% significance level, then we MUST also reject it at 1% level **False**. d)A type II Error cannot be committed when **i) Hois true**. e)Probability of Type I Error is called **significance level**. f)If the confidence level decreases, then the confidence interval becomes wider **False**. g)Let Z be the standard normal r.v.. Find P(Z = 1.28)=**0**. h)The sampling distribution of the sample mean has the mean = **μ**. a)only if n≥ 30 **b) always** i)T-distribution and standard normal distribution have the same variance **False. **2)The number of defective items produced by a machine for each of the last 23 days are as follows: **(Valentina are the numbers in point b)**a)Construct a cumulative frequency distribution (use class width of 5)**Smallest = 5, Largest = 29**. b)Calculate Q1, Q3and IQR for this data set. **Ordered data **5 5 6 7 8 **9** 10 12 14 14 15 **15** 17 17 19 19 20 **22** 23 23 27 28 29. **Q1=9 Q2=15 Q3=22. IQR=Q3-Q1=22-9=13.** 3)An insurance company survey indicates that 42% of their customers have a speed ticket, 54% are female and 24% are females and have a speed ticket.a)What proportion of customers have a speed ticket or are females? Speed ticket(S)Not ticket(N)Female (F)Male(M) **P(S or F)=P(S U F)=0.18+0.24+0.30=0.72(or using the formula,=0.42+0.54–0.24=0.72**). b)What proportion of the females have a speed ticket? **P(S|F)=0.24/0.54=0.44 **4)The voting rate in a large university is 70%. 15 students were randomly selected from that university and asked if they voted or not.a)Calculate the mean number of students that voted from the 15 selected. **This is a Binomial r.v., therefore, mean=μ=E(X)=np=15(0.7)=10.5**. b)Calculate the standard deviation of the number of students that voted from the 15 selected. **σ=√(variance)=√(np(1–p))= √(15(0.7)(1–0.7))=√(3.15)=1.7748**. 5)The average cholesterol content of a certain brand of eggs is 215mg and the standard deviation is 15 mg. Assume the cholesterol content is normally distributed.a)Find the probability that a randomly selected egg has the cholesterol content more than 222 mg.**P(X>222)=P(Z>(222 –215)/15)=P(Z>0.47)=1–0.6808=0.319** b)Calculate the 15thpercentile of the cholesterol content for this brand of eggs. **From the Z-table, the 15 ^{th} percentile is z=–1.04 –>1.04=(x–215)/15 –> x=215–1.04(15)=199.4 **c)If 25 eggs are selected, find the probability that the sample mean cholesterol content is less than 218 mg

**. P(X<218)=P(Z<(218 –215)/(15/√(25))= P(Z<1) =0.841.**6)A department store finds that in a random sample of 80 customers, 35%of them had browsed its website prior visiting the store. Find a 98% confidence interval for the proportion of all customers that browse the store’s website prior visiting the store.

**P±z**. 7)A sample of 36 pairs of tennis shoes was used to test if the average cost is greater than $69.95. Assuming the population standard deviation of cost is $3.00, is there enough statistical evidence at 1% significance level to infer that? The Test Statistic was calculated to be 0.6. Calculate p-value only (do NOT write all the steps!).

_{α/2 }√{[p(1-p)]/n}= 0.35±2.33 √{[0.35(1-0.35)]/80}=0.35±0.1243**This is a right-tail test and sigma is known, so it is a Z-test p-value=P(Z>T.S)=P(Z>0.6)=1–0.7257=0.2743.**8)Accordingly to a study several years ago, the average wireless phone user earns $62,600 per year. A researcher wants to test if the average is lower now, at 1% significance level. He randomly selects 16 wireless phone users and finds their average annual salary to be $58,974 and the standard deviation $7,810. Assume annual salaries are normally distributed. a)Hypotheses and Test Statistic.

**H0: μ≥62600 (or μ=62600) vs HA: μ<62600 (left-tail). The standard deviation of the population is unknown, so we use standard deviation of the sample, thus this is a T-test T.S = t**b)Rejection Region and Decision

_{15}= (x-μ_{0})/(s/√n)= (58974-62600)/(7810/√16)=-1.8571.**the critical value = −t**c)Conclusion(use a full sentence pertaining to this particular problem)

_{α,df}= −t_{0.01,15}= −2.602, so rejection region is T<−2.602T.S. is not in the R.R.àDo not reject Ho**There is not enough statistical evidence at 1% significance level to infer that average wireless phone user earns less than $62,600 per year**. 9)A sample of ball bearings produced by a machine was selected and the mean diameter was found to be .55 inch. Use the appropriate Excel output to estimate with 99% confidence the mean diameter of all ball bearings produced. The diameter it is known to have a normal distribution with a population standard deviation of 0.0127. Cross out the wrong Excel outputs and write your answer in a full sentence pertaining to this problem.

**We are 99% confident that mean diameter of all ball bearings produced is between 0.5429 and 0.5579 inches.**10)A market researcher wants to prove that, on average, Raymondville

**(1)**residents dine out more often than Rosenberg

**(2)**residents. He selects a random sample of 64 families from each suburb and asked them how many times they dined out last month. Excel was used to run the test and p-value was .00087. Write the hypotheses and your conclusion at 5% significance level

**. H0: μ**11)An engineer wants to determine the effectiveness of a safety program. He collects annual loss of hours due to accidents in 12 plants before and after the program was put into operation. Assuming the difference to be normally distributed, he wants to test the effectiveness of the program at 1% significance level. Use the Excel output to write the Hypotheses and your final conclusion. P(T≤ one tail) 0.00666

_{1}≤ μ_{2}(μ_{1}–μ_{2}≤ 0 or μ_{1}–μ_{2}= 0) vs. HA: μ_{1}> μ_{2}(μ_{1}–μ_{2}> 0). Since 0.00087 < 0.05, there is enough evidence at 5% significance level to infer that on average, Raymondville residents dine out more often than Rosenberg residents.**H0: μ**12)To test if average cholesterol level is significantly higher for people in the south

_{D}= 0vs HA: μ_{D}> 0, since p-value = 0.0066 < 0.01, there is enough evidence at 1% significance level to infer that the program was effective.**(1)**than for people in the north

**(2),**which of the following test should be used? (circle one). Write the hypotheses for the test you choose. •T-test Paired Two samples for means–Matched -Pairs Samples

**•T-test Two-samples for means –Independent Samples H0: μ**•Z-test Two samples for proportions–Independent Samples. 13) A bank officer wants to estimate the average total monthly deposits per customer at the bank. How large a sample should be selected to be within $100 from the actual average, with 90% confidence? He assumes the standard deviation of total monthly deposits for all customers is about $1,000.

_{1 }≤ μ_{2}(μ_{1}= μ_{2}) vs. HA: μ_{1}> μ_{2}**n= ((z**

_{α/2}*σ)/D)^{2}= ((1.645×1000)/100) =270.603à271 customers.