**Review
of Statistical Inference from STA 2023**

** **

Confidence Intervals and Significance Tests
for:

·
one mean

·
matched paired differences

· difference of two independent means

· one proportion

· difference of two independent proportions

** **

**Chapter 7: CI and
Significance Tests for means using the t distribution**

**Why
do we use the t table?**

What assumptions do we need to make?
How do we check them?

1. Random Samples

2. Population is Normal

Find
Formulas on tables.

§ CI:

§
Sig. Test: H_{o}:_{ } H_{a}:

_{ }

TS:

p-value:

**Read Minitab Output **

** **

** **

**Interpret Results **

CI does NOT include # from H_{o}
→ Supports H_{a}

p-value small → Supports H_{a}

**Chapter 8 : CI
and Sig. Tests for p and p _{1}-p_{2 }using the z**

** **

**Why do we use the z
table?**

**What assumptions do
we need to make? **

1. Random Samples

2. Check specific assumption for that test

§ One Sample Proportion (Confidence Interval)

Sample size must be at least equal to 5.

§ One Sample Proportion (Significance Test)

**Must have at least
ten successes and ten failures.**

Both sample sizes must be at least equal to 10.

§ Two Sample Proportions (Significance Test)

The number of successes and the number of failures must be at least 5 in each of the samples.

§
**EASIEST – Use the most restrictive one for all cases
and you will be ok.**

**Which estimator of
p do we use? **

·
One Sample Proportion (Confidence Interval) _{}

·
One Sample Proportion (Significance Test) _{}

·
Two Sample Proportion (Confidence Interval)** **_{} _{}

·
Two Sample Proportions (Significance Test) _{}

** **

**Examples: For each of the
following examples, we will first determine which kind of problem it is. The Minitab results for each problem are
given in the last page. **

1. Suppose that in a random sample of 50 babies
conceived by an *in vitro *fertilization
process (test tube babies), 35 are girls. Does this sample evidence show
convincingly that the* in vitro*
process favors the female gender? Explain.

2. Four sets of identical twins were selected at random. One
child was selected at random from each pair to form an "experimental
group"- these four children were sent to school. The other four
children were kept at home as a control group. At the end of the school
year the IQ scores presented below were obtained. Does this evidence
justify the conclusion that lack of school experience has a depressing effect
on IQ scores?

Experimental
Control

__
Pair
Group
Group__

A
110 111

B
125
120

C
139
128

D
142
135

3. A restaurant has recently started using a dessert
cart that is pushed around the eating area for diners to view and, it is hoped,
be unable to resist. On the first ten evenings the cart
was used, the average expenditure per night for
desserts was $130 compared to an average of $110 prior to the use of the
cart. The standard deviation for the ten evenings was $65. Does it
look like the cart will improve sales significantly?

4. A social scientist believes that the fraction of Republicans in favor of the death penalty is greater than the fraction of Democrats in favor of the death penalty. She asked random samples of 200 Republicans and 200 Democrats and found 46 Republicans and 34 Democrats in favor of it. Do these data support the social scientist's belief?

5. The effect of alcohol consumption on the body appears to be much
greater at high altitudes than at sea level. To test this theory, a
scientist randomly selects 12 subjects and randomly divides them into two
groups of six each. One group is transported to an altitude of 12,000
feet, where each subject ingests a drink containing 100 cc of alcohol.
The second group receives the same drink at sea level. After two hours,
the amount of alcohol in the blood (grams per 100 cc) for each subject is
measured. The data are shown in the table. Do the data provide
sufficient evidence to support the theory that retention of alcohol in the
blood is greater at high altitudes?

Sea Level |
12,000 Feet |

0.07 |
0.13 |

0.10 |
0.17 |

0.09 |
0.15 |

0.12 |
0.14 |

0.09 |
0.10 |

0.13 |
0.14 |

**Test
and Confidence Interval for One Proportion**

**Test of p = 0.5 vs p > 0.5**

Exact

X N
Sample p 95.0 %
CI P-Value

35 50 0.700000 (0.553918, 0.821382) 0.003

Test and
CI for One Proportion with Wilson’s Estimate

Sample X
N Sample p 95.0% CI

1 37 54 0.685185 (0.544476, 0.804761)

** **

**Paired
T-Test and Confidence Interval**

**Paired T for exp - control**

** **
N Mean StDev SE
Mean

exp 4
129.00 14.67 7.34

control 4
123.50 10.34 5.17

Difference 4
5.50 5.00 2.50

95% CI for mean difference: (-2.46, 13.46)

T-Test of mean difference = 0 (vs >0): T-Value =
2.20 P-Value = 0.058

**T-Test of the Mean**

**Test of mu = 110.0 vs mu > 110.0**

Variable N Mean StDev
SE Mean T P

dessert 10 130.0 65.3
20.6 0.97 0.18

**T
Confidence Intervals**

Variable N Mean StDev SE
Mean 95.0 % CI

dessert 10 130.0 65.3 20.6
(83.3,176.7)

**Test
and Confidence Interval for Two Proportions**

Sample
X N Sample p

1 46 200 0.230000

2 34 200 0.170000

Estimate for p(1) - p(2): 0.06

95% CI for p(1) - p(2): (-0.0181778, 0.138178)

Test for p(1) - p(2) = 0 (vs > 0): Z = 1.50 P-Value = 0.066

**Test and
CI for Two Proportions with Wilson’s Estimate**

Sample X
N Sample p

1 47 202 0.232673

2 35 202 0.173267

95%
CI for p(1) - p(2): (-0.0188205,
0.137632)

**Two
Sample T-Test and Confidence Interval**

**Two sample T for sea level vs 12000ft**

N Mean StDev SE Mean

sea leve 6 0.1000 0.0219 0.0089

12000ft 6 0.1383 0.0232 0.0095

95% CI for mu sea leve - mu 12000ft:( -0.0678,
-0.0089)

T-Test mu sea leve = mu 12000ft (vs <): T = -2.94 P = 0.0082 DF= 9