# ded

1)Fill in the blanks or circle one answer a)UNF has 16,719
students of which 60% are female. A statistician wants to estimate the mean
number of credit hours per semester for students at UNF. He plans to randomly
select 60 female and 40 male students. What sampling plan has he chosen? **Stratified** b)A statistician wants to
estimate the mean number of credit hours per semester for students at UNF. He
plans to randomly select 5 classrooms and interview all the students in there.
What sampling plan has he chosen? **Cluster**
c)A statistician wants to estimate the mean number of credit hours per semester
for students at UNF. He randomly selected 150 numbers from 1 to 16,719andthen
he selected those students from a student list provided by admission
department. What sampling plan has he chosen? **Simple random** d)The mean of X = **μ** and standard deviation of X is
= **σ****/√n** e)If the
confidence level decreases than the confidence interval becomes wider **False** f)If the sample size decreases
than the confidence interval becomes shorter **True** g) Notation for confidence level is **1 –α** h)90% C.I. for mean is (137, 143). What is the error of this
estimate? **3** i)90% C.I. for mean is
(137, 143). What is the point estimate? **140**
j)90% C.I. for mean is (137, 143). What is the confidence level of this
estimate? **90%** k)If the confidence
level decreases than the confidence interval becomes wider **False** l)If the sample size decreases than the confidence interval
becomes shorter **False** m)Notation for
significance level is **α **n)If we
reject Ho at 5% significance level, then we must also reject it at the 1%
level. **False** o)If we do not reject
Ho at 5% significance level, then we must also not rejecting it at the 1%
level. **True** p)If we reject Ho at 5%
significance level, then we must also reject it at the 10% level. **True** q)A Type II Error cannot be
committed when: **Ho is true** r)A Type
I Error cannot be committed when: **Ho is
false** s)A Type II Error can be committed when: **Ho is false** t)A Type I Error can be committed when: **Ho is true**. u)Probability of Type I
Error is called **Significance level** v)Power
of the test is **1–P(Type II error)** w)Probability
of Type II error is denoted by **β** x)Probability
of Type I error is denoted by **α** y)Notation
for power of the test is **1–β** z)In a
hypothesis test if we increase α than β also increases **False. **2)The grades of all the exams of one professor has mean µ=75
and standard deviation σ=4. a)What is the distribution of sample means of
samples of 36 exams? What about its mean and its standard deviation? **Shape normal, because n=36 > 30, mean=75,
st dev=4/6. **b)What is the distribution of sample means of samples of 25
exams? What about its mean and its standard deviation? **Shape unknown because n=25<30 and population distribution is
unknown, mean=75, st dev=4/5. **3)The weekly output of a steel mill is
normally distributed with mean of 130 tons and standard deviation of 20 tons.
a)What is the probability that the mill will produce less than 135 tons next
week?** P(X<135)=P(****????<(135−130)****/20
)=P(Z<0.25)=0.5987. **b)What is probability that the mean weekly output of
9 randomly selected weeks is less than 135 tons? **P(****????<135)=P(****????<135−130****/(20****/√9)=P(Z<
0.75)=0.7734. **c)What is probability that the mean weekly output of 52
randomly selected weeks is less than 135 tons? **P(****???? <
135) = P(****????<135−130****/(20****/√52)
= P(Z < 1.80) = 0.9641**. 4)60% of UNF students are female. Suppose you
decide to randomly select 15 UNF students. a)What is the expected value of the
proportion of females in this sample? **E(****????) = p = 0.6 **b)What
is the standard deviation of the proportion of females in this sample? **√{(0.6)(1−0.6)****/15}=0.1265 **c)Calculate
the probability that proportion in the sample is less than 35% **P(****????
< 0.35) = P(****????
< (0.35−0.6)****/0.1265 ) = P(Z < – 1.98) =
0.0239 **d) Calculate the probability that proportion in the sample is more
than 38%. **P(****???? > 0.38)= P(****???? > (0.38−0.6)****/0.1265)=P(Z>-1.74)
=1– 0.0409=0.9591** e)Calculate the probability that proportion in the sample
is between 38% and 41%. **P((0.38−0.6)****/0.1265<****????<(0.418−0.6)****/0.1265)=P(–
1.74<Z<–1.5)=0.0259. **5)The temperature readings for 40 randomly
selected winter days in Grand Rapids, MI have a mean of 5.5 F degrees. Assuming
temperatures have a population variance of 2.1, determine the 95% confidence
interval estimate for the mean winter temperature. Write your answer in a full
sentence pertaining to this particular problem. **5.5±****???? _{.025}
(√2.1**

**/√40)=5.5±1.96(√2.1**

**/√40)=5.5±0.4491=5.05091 to 5.9491.**

**We are 95% sure that average winter temperature in Grand Rapids is between 5.05F and 5.95F.**6) The manager of a newly opened Target store wants to estimate the average expenditure of his customers with 99% confidence level. How many customers he should select to estimate the average with a margin of error of $10, knowing that an estimated standard deviation is $18.

**????**

**=((2.575*18)**

**/10)**7)Ages of employees in a company are normally distributed. A sample of ages of 16 employees yielded a mean of 32 and a standard deviation of 3. a)Can we use the Z-interval, T-interval or neither to estimate µ? Why?

^{2}=21.48, therefore the manager should select 22 customers.**T-interval, because standard deviation of the population is unknown and population is normal.**b)What is the point estimate?

**32**c)Estimate the average age of all employees with 98% confidence.

**32±**

**????.**

_{01,15}(3**/√16) =32±**

_{2.602}(3**/√16)= 32±1.9515=30.05 to 33.95yrs.**d) Interpret the confidence interval found above.

**We are 98% confident that average age of work for all employees in this company is between 30.05 years and 33.95 years.**e)What is the margin of error of this estimate?

**E = 1.9515**f)The math tests scores in a certain university are normally distributed. A sample of 49 tests yielded a mean of 80 and a standard deviation of 5. Estimate the mean test score at this university with 90% confidence. Interpret.

**80±**

**????.**

_{05,48}(5**/√49) = 80±1.677 (5**

**/√49) = 80 ± 1.979 = 78.8 to 81.2. We are 90% confident that average math score in this university is between 78.8 and 81.2**8)Calculate the sample size needed to estimate the percentage of defective items from a production line at Vistakon to within 2% with 95% confidence level.

**We do not have any info about p, therefore we use 0.5;**

**????= (0.5)*(1 − 0.5)*(1.96/0.02)**9)Nielsen Media Research uses samples of 5,000 households to rank TV shows. Suppose Nielsen reports that NFL Monday Night Football had 35% of the TV audience. What is 99% confidence interval for this parameter?

^{2}= 2,401 items**0.35±**

**????.**10) The manager of a manufacturer found 11 defective items in a sample of 121 items. a)What is the point estimate for the proportion of defective items in this manufacturer?

_{005 }√{(0.35)*(1−0.35)/5000}= 0.35 ± 2.575*(. 0067) = 0.35 ± 0.0174 = 33.3% to 36.7%**????**

**= 11/121 = 0.0909**b) Can we use a Z-interval to estimate proportion of defective items in this manufacturer?

**Yes,because 121*(0.0909)=11>5 and 121(1-0.0909)= 110>5**c)Estimate the proportion of defective items in this manufacturer with 95% confidence.

**0.0909 ±**

**????.**d)Interpret the confidence interval found above.

_{025 }√{[0.0909*(1−0.0909)]/121} = 0.0909 ± 1.96*(0.2875)= 0.0909±0.0512 = 0.04 to 0.14**We are 95% confident that proportion of defective items at this manufacturer is between 4% and 14%**e)What is the margin of error of this estimate?

**E = 0.0512**f)The manager is not satisfied with the length of the interval obtained. How many items the manager should select to estimate the proportion to within 1% with the same confidence?

**We do have an estimate of p as 0.0909;**

**???? =0.0909*(1 − 0.0909)*(1.96/0.01)**11)Chicken Delight claims that 90% of its orders are delivered within 10 minutes of the order time. A researcher wants to test if less than 90% of the orders are delivered “on time”. a)Write the Hypotheses of this test.

^{2 }=3174.87 to 3,175 items**Ho: p ≥ .9 vs HA: p < .9**b)Is this a right-tail, left-tail or two-tail test?

**left-tail**12) The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 96.4% of the people who have that disease. However, it erroneously gives a positive reaction in 2.3% of the people who do not have the disease. Consider the null hypothesis “the individual does not have the disease” to answer the following questions. a) What is the probability of Type I error?

**Ho: the individual does not have the disease. HA: the individual has the disease. Type I error is to conclude that the individual has the disease (positive reaction), when in fact the individual does not have the disease, therefore its probability is α = 0.023.**b)What is the probability of Type II error

**? Type II error is to conclude that the individual does not have the disease (negative reaction), when in fact the individual has the disease, therefore its probability is β = 1 – 0.964 = 0.036.**13) A sample of 30 pairs of tennis shoes has an average cost of $70.25, is there enough evidence at 1% significance level, to infer that average cost is greater than $69.95? Assume population standard deviation is $1.87. a)Write the Hypotheses

**Ho: µ ≤ 69.95 vs HA: µ > 69.95**b)Is this a right-tail, left-tail or two-tail test?

**right-tail**c) Calculate the Test Statistic (T.S.).

**TS= (70..25−69.95)/[1.87/√30] = 0.8787**d)Calculate p-value.

**P(Z > 0.88)= 1– 0.8106= 0.1894**e)Do we reject or we do not reject Ho? Why

**? P-value = 0.1894>α= 0.01, therefore we do not reject Ho.**f)Write the conclusion pertaining to this problem.

**There is NOT enough statistical evidence at 1% significance level to infer that average cost for tennis shoes is above $69.95.**