Download Stat and Data Analysis Name: One Sample t-interval

Stat and Data Analysis
One Sample t-interval and t-test
Name: ____________________
1. A company has set a goal of developing a battery that lasts over 5 hours (300 minutes) in
continuous use. In a first test of a SRS of 12 of these batteries measured the following lifespans (in minutes): 321,295, 332, 351, 281, 336, 311, 253, 270, 326, 311, and 288.
a. Is there evidence that the company has met its goal? Assume that the life-spans of the
batteries is approximately Normal. Create an appropriate test.
Conditions:
1. SRS
2. n > 30
3. pop > 10n
Stated
12 < 30 but Normal Distribution
120 there are more than 120 batteries
x = 306.25 s = 29.31
n = 12 df = 11
HO: μ = 300
HA: μ > 300
306.25 − 300
= 0.7387
29.31
12
P-Value = P(t11 > 0.7387) = 0.2378
Since the P-Value is greater than alpha(0.2378 > 0.05) we fail to reject the null
hypothesis. There is not enough evidence that the mean life-span of the batteries
is more than 300 minutes.
t11 =
b. Create a 98% confidence interval for the mean lifespan of this type of battery.
df = 11
∗
t11
= 2.718
29.31
306.25 ± 2.718
12
CI = 306.25 ± 22.30
( 283.25,329.25 )
We are 98% confident that the true mean life-span of the batteries lies between 283.25
to 329.25 minutes.
2. Consumer Reports tested a SRS of 14 brands of vanilla yogurt and found the following
numbers of calories per serving:
160 200 220 230 120 180 140
130 170 190
80 120 100 170
a. A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What
does this evidence indicate? Use your confidence interval to test an appropriate
hypothesis and state your conclusion.
Conditions:
1. SRS
2. pop > 10n
3. n > 30
HO: μ = 120
HA: μ ≠ 120
Stated
10(14) = 140 There are more than 140 yogurts
14 < 30 but Normal Distribution
x = 157.86 s = 44.75
n = 14 df = 13
157.86 − 120
= 3.1656
44.75
14
P-Value = 2•P(t13 > 3.1656) = 0.0074
Since the P-Value is less than alpha(0.0074 < 0.01) we reject the null hypothesis.
There is significant evidence that the mean calories from a serving of vanilla
yogurt is not 120 calories.
t13 =
b. Create a 95% confidence interval for the average calorie content of vanilla yogurt.
(Assume the number of calories is unimodal and symmetric)
x = 157.86 s = 44.75
n = 14 df = 13
∗
= 2.160
t13
CI = 157.86 ± 2.160
44.75
14
CI = 157.86 ± 25.83
(132.03,183.69)
We are 95% confident that the true mean calorie content of vanilla yogurt lies between
132.03 and 183.69 calories.
3. How much time do school-age children spend helping with housework? A random sample of
26 girls in two-parent families where both parents work full-time found a mean of 14.0
minutes per weekday with a standard deviation of 8.6 minutes. A histogram of the data
showed a unimodal and approximately symmetric distribution. Construct and interpret a
98% confidence interval for the mean number of minutes spent doing housework.
Conditions:
1. SRS
2. n > 30
3. pop > 10n
- Stated
- n = 26X Data is unimodal and symmetric
- 10(26) = 260
∗
t 25
= 2.485
x = 14.0 s = 8.6
8.6
n = 26 df = 25
CI : 14.0 ± 2.485
26
CI : 14.0 ± 4.19
CI : ( 9.81,18.19 )
We are 98% confident that the true mean number of minutes spent doing housework
lies between 9.81 and 18.19 minutes.