Download Solutions a) A confidence interval needs a two tailed probability. 90

Solutions
a) A confidence interval needs a two tailed probability.
90% confidence level means a total of 0.10 in the tails of the probability distribution.
Go to df = 17 and find the critical value tc = 1.740
b) A confidence interval needs a two tailed probability.
98% confidence level means a total of 0.02 in the tails of the probability distribution.
ff =88 is not in the table. Go to closest one which is df = 100 and find the critical value
tc = 2.364
c) In the row with df = 4, find the value t = 2.09 in the main body of the table.
The closest value is t = 2.132. This corresponds to a probability of 0.10 in both tails, which means
we have a probability of 0.05 in one of the tails, t > 2.04
With the TI83/84:
Click on 2nd Vars. This will take you to the Distribution Menu.
Scroll down and choose tcdf.
The t-distribution required: tcdf(lower value, upper value, degrees of freedom.
Here we want the probability that t > 2.04 so we type in:
tcdf(2.04, 100000, 4) and we get .05546
d) In the row with df = 22, find the value t = 1.78 in the main body of the table.
The closest value is t = 1.717. This corresponds to a probability of 0.10 in both tails. This is the
answer we want, since |t| > 1.78 means the probability in both tails.
With the TI83/84:
Click on 2nd Vars. This will take you to the Distribution Menu.
Scroll down and choose tcdf.
The t-distribution required: tcdf(lower value, upper value, degrees of freedom.
Here we want the probability that |t| > 1.78 so we type in:
tcdf(1.78, 100000, 22) and we get .04444. This gives the probability in the upper tail only.
|t| > 1.78 means the probability in BOTH tails, so to get the answer we multiply by 2:
2 x .04444 = .08888
Solutions
a) We are 95% confident that the true mean of pulse rates for adults is between 70.9 to
74.5 beats per minute.
b) The formula for the confidence interval for the mean is:
̅
̅
The confidence interval can also be written as: ̅
√
,
df = n-1
̅
̅
Comparing this with the results:
̅
̅
Add the two equations:
̅
Plug this into one of the equations:
beats per minute
c) An interval with 99% confidence would have a larger margin of error, so it would be
larger.
Solution:
a) The formula for the confidence interval for the mean is:
̅
Here, ̅
̅
√
,
df = n-1
n = 40, so df = n – 1 = 39
Next, we need to find the critical t-value. To do this we use the t-distribution table.
Confidence intervals need two-tailed probability. Since we need 95% confidence, this
means that .05 of the probability is in the two tails.
We look in the column that starts with 0.05 and go down to the row with df = 39. Since
there is no row for df = 39, use df = 40.
The t-value = 2.021 (see table on the next page).
√
b) Assumptions: normal population, random sample
It is reasonable to assume that the weight of hot dogs is distributed normally.
c) We are 95% confident that the mean amount of sodium in low-sodium hot dogs falls
between 298.5 mg to 321.5 mg.
Solution:
a) The formula for the confidence interval for the mean is:
̅
Here, ̅
̅
√
,
df = n-1
n = 23, so df = n – 1 = 22
Next, we need to find the critical t-value. To do this we use the t-distribution table.
Confidence intervals need two-tailed probability. Since we need 95% confidence, this
means that .05 of the probability is in the two tails.
We look in the column that starts with 0.05 and go down to the row with df = 22.
The t-value = 2.074 (see table on the next page).
√
The result above is the confidence interval for the values when 299,000 was subtracted
from the actual speed of light. To get the confidence interval for the actual speed of
light, we add 299,000 to the lower and upper values:
b) We are 95% confident that the mean speed of light falls between 299,709.9 km/sec and
2999,802.54 km/sec
Solution:
a) Yes, a larger sample size would give a better estimate (smaller margin of error.)
me is smaller because we are dividing by a larger value of n.
Also, the critical t-value gets smaller as we increase the value of n.
b)
√
√
mg of sodium
c) The formula for the confidence interval for the mean is:
̅
Here, ̅
̅
√
,
df = n-1
n = 60, so df = n – 1 = 59
Next, we need to find the critical t-value. To do this we use the t-distribution table.
Confidence intervals need two-tailed probability. Since we need 95% confidence, this
means that .05 of the probability is in the two tails.
We look in the column that starts with 0.05 and go down to the row with df = 59. Since
there is no row for df = 39, use df = 60.
The t-value = 2.000
√
We are 95% confident that the mean amount of sodium in low-sodium hot dogs falls
between 309.74 mg to 326.26 mg.
d) 30% less than 465 mg means that low sodium hot dogs should have 70% of the sodium level of
regular hot dogs: .7(465) = 325.5 mg
Since this value is inside the confidence interval, it is possible that these hot dogs should not be
labeled are “reduced sodium”.
Solution:
a) It is not specified if it is a random sample.
The Near Normal Condition seems reasonable from a Normal probability plot. The
histogram is nearly uniform, with no outliers or skewedness.
b) The formula for the confidence interval for the mean is:
̅
̅
√
,
df = n-1
We need to calculate the mean and standard deviation of the sample, either by hand or
with the TI calculator.
With the TI, go to Stat Edit and type the 6 values in L1. Then go to Stat Calc and then
choose 1 Var Stats, hit Enter and Enter again.
This gives the following:
Here, ̅
n = 6, so df = n – 1 = 5
c) Next, we need to find the critical t-value. To do this we use the t-distribution table.
Confidence intervals need two-tailed probability. Since we need 95% confidence, this
means that .05 of the probability is in the two tails.
We look in the column that starts with 0.05 and go down to the row with df = 5.
The t-value = 2.571
√
d) We are 95% confident that the mean weight of the content of Ruffles bags is between
28.36 g and 29.21 grams.
e) Since all values are larger than 28.3 grams, it appears that the company is playing it safe
and they are overfilling the bags slightly.
a) Find the means and standard deviation using the TI83/84:
With the TI, go to Stat Edit and type the 6 values in L1. Then go to Stat Calc and then
choose 1 Var Stats, hit Enter and Enter again.
This gives the following: ̅
Here is the plot of the z-scores vs the time values:
The conditions for inference seem to be satisfied because the normal probability plot is linear.
We have a possible outlier at 93.8 seconds.
b) Null hypothesis: Ho: μ = 60
Alternative hypothesis: Ha: μ ≠ 60
̅
Test statistic:
√
√
P-value = P(|t| > 2.63)
With the table, find the value closest to 2.63 in the row with df = 20. This value is 2.528.
Go to the top of the column, to the two-tailed probability to get p-value = 0.02
With the TI83/84:
Click on 2nd Vars. This will take you to the Distribution Menu.
Scroll down and choose tcdf.
The t-distribution required: tcdf(lower value, upper value, degrees of freedom.
Here we want the probability that |t| > 2.63 so we type in:
tcdf(2.63, 100000, 20) and we get .008. This gives the probability in the upper tail only.
|t| > 2.63 means the probability in BOTH tails, so to get the answer we multiply by 2:
2 x .008 = .016
The p-value is small. With this p-value we might conclude that the mean completion time for the
maze is not 60 seconds.
c) Remove the outlier. Calculate the new mean and standard deviation using the TI83/84.
̅
Test statistic:
̅
√
√
P-value = P(|t| > 4.46)
With the table, find the value closest to 4.46 in the row with df = 19. There is no such value, pvalue is approximately 0.
With the TI83/84:
Click on 2nd Vars. This will take you to the Distribution Menu.
Scroll down and choose tcdf.
The t-distribution required: tcdf(lower value, upper value, degrees of freedom.
Here we want the probability that |t| > 4.46 so we type in:
tcdf(4.46, 100000, 19) and we get .000134. This gives the probability in the upper tail only.
|t| > 4.46means the probability in BOTH tails, so to get the answer we multiply by 2:
2 x .000134 = .00027
Because the p-value is so small, we reject Ho. Without the outlier, we see strong evidence that
the average completion time for the maze is less than 60 seconds. The outlier did not change
the conclusion.
d) The maze does not meet the “one-minute average” requirement. Both tests rejected a null
hypothesis of mean of 60 seconds.