Question

1. The service manager of an appliance store decides to raffle a 50 in. flat screen television. She prepares a list of 500 regular customers from a list of recent purchases. Each customer is assigned a number from 200 to 699. The manager decides to sample 20 customers and enter them in the grand prize drawing. Use the LAST three digits in the Random Number Table starting at row 30, columnl, and reading across the list. List the 20 customer numbers. Only list the last three digits. 2. An automobile manufacturer claims that their leading compact car averages 36 mpg in the city. The population standard deviation is 8 mpg for city driving. Suppose a city police department purchases 64 cars from this auto manufacturer. If these cars were driven exclusively under city conditions and averaged 34 mpg, can one argue that the manufacturer's claim is too high at 1%. a. State the null and alternative hypothesis b. Determine and draw the hypothesis test model (Label everything) C. Determine the critical value(s) and test statistic d. Determine the conclusion 3. Presently, the mean life expectancy of a rare strain of bacteria is 12 hours. A scientist claims that she has developed a medium and therefore disagrees with this hypothesis that it is not 12 hours. The scientist tests 25 cultures and finds that they have a mean life of 13.25 hours with a sample standard eviation = 2 hour. Do these results show that the medium is effective in increasing the bacteria's life expectancy? (Use <= 5%) a. State the null and alternative hypothesis b. Determine and draw the hypothesis test model (Label everything) c. Determine the critical value(s) and test statistic d. Determine the conclusion 4. The probability that Casey, a softball player gets a hit during a single time at bat is 3/10. If she goes to bat 6 times in her next game, determine the probability to four decimal places a. That she gets at least 3 hits in the game. b. That she gets at most 4 hits in the game c. That she does not get a hit in the game. d. That she gets exactly 3 hits in the game
show work

Answer 1

2.

Let u denotes the population mean. Hypotheses : To test null hypothesis H. :u= 36 against alternative hypothesis H1 :p < 36 Let i denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, T = 34.00,02 = 64.0000,0 = 8.00, n=64 The test statistic can be written as: Vn(ē-36) which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if Zobs < -20.01 = -2.326348 0.4 0.3 density 0.2 0.1 - 0.0- -4 -2 N Now, Value of test statistic: The value of the test statistic Zobs V64 (34.00 – 36) = -2.000000 ~ -2.00 8.00 and the critical value is critical = -2.326348 -2.33 P-value = P(Z < Zobs) = P(Z < -2.000000) = 0.022750 – 0.0228 Conclusion : Since p-value > 0.01 and Zobs Zcritical = -2.326348, so we fail to reject H, at 0.01 level of significance Hence, we can conclude that population mean is not significantly less than 36

3.

There is sufficient evidence to support the claim that the medium is effective in increasing the bacteria's life expectancy.

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho pe=12 against alternative hypothesis H : > 12 Let e denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, = 13.250, 82 = 4.000000, s = 2.00, n=25 The test statistic can be written as: /(– 12) te s which under H, follows at distribution with n-1 df. Decision rule / Rejection region : We reject H, at 0.05 level of significance if P-value < 0.05 or if tobs > t0.05,n-1 = 1.710882 0.4 0.3 0.2 A 0.1 - 0.0 - -2 0 t Now, Value of the test statistic : The value of the test statistic tobs = V25 (13.2500 - 12) 2.0000 = 3.125000 ~ 3.12 associated degrees of freedom = 24 The critical value = 1.710882 1.71 P-value = P(t24 > 3.125000) =1- P(t24 <3.125000) = 1-0.997699 = 0.002301 0.0023 Conclusion : Since p-value < 0.05 and tobs > tcritical = 1.710882, so we reject H, at 0.05 level of significance Hence, we can conclude that population mean is significantly greater than 12

4.

Let X denotes the number of times that she hits among 6 times.

Here,

X ~ Binomial(n = 6, p = 3/10 = 0.3)

The probability mass function of X is

P(X = z) = *0.3" * (1 – 0.3)6-1, x = 0,1,2,3,4,5,6

a)

P(X23) = 1 – P(X < 3) = 1 – P(X < 2) = 1 - Σ(3) - * 0.3" * (1 0.3)6-r =1 – 0.74431 = 0.25569

b)

$P(X \leq 4) =\sum_{x=0}^{4} \binom{6}{x}*0.3^x*(1-0.3)^{6-x} = 0.989065 \approx 0.9891$

c)

$P(X =0) =\binom{6}{0}*0.3^0*(1-0.3)^{6-0} = 0.117649 \approx 0.1176$

d)

P(X =3) = (9) ** *0.33* (1 -0.3) 6-3 = 0.18522