Blood pressure of 60 year old women with glaucoma

The confidence interval for the population mean is given by:

N

200

Mean

140

Standard Deviation

25

SE Mean

1.77

                        Confidence interval for population mean is (136.51, 143.49)

           There is 95% confidence that true mean will be between 136.51 and 143.49

The confidence interval for the sample mean is given by:

N

100

Mean

140

Standard Deviation

25

SE Mean

2.5

The 95% confidence interval for sample mean is (135.04, 144.96)

There is 95% confidence that true mean will be between 135.04 and 144.96

The interval gets wider when the sample size is decreased by 100 from (136.51, 143.49) to (135.04, 144.96).

Reason:  Because reducing the sample size (n) increases the standard error of the mean from 1.77 to 2.5 hence the confidence interval gets wider.

  

Paired t-test/Confidence interval for matched samples

Then the 95% confidence interval   for the mean difference  is given by:

Difference

Sample size

107

Mean

2.1

Standard Deviation

3.1

SE Mean

0.3

                  The 95% confidence interval for mean difference is (1.506, 2.694)

There is 95% confidence that true mean is located in the interval of (1.506, 2.694), it means that PTCA is effective in increasing exercise duration. The problem of constructing confidence interval for the difference of the means from two populations with unknown variances is difficult. 

The 95% confidence for the mean difference is given by:

Vitamin E

Placebo

Sample size

81

90

Mean

27

24

Standard Deviation

6.9

6.20

SE Mean

0.77

0.65

The  95% confidence interval for the sample difference is (1.02, 4.99)

Two sample t test

The problem of unknown variances of the two groups.

There is 95% certainty that true mean is located between 1.02 and 4.99, thus the conclusion is Vitamin E increases cognitive ability among elderly women.

The test hypothesis is

Ho: There is a difference in cooling constants between freshly killed and reheated corpse.  Vs Ha: There is  no difference in cooling constants between freshly killed and reheated corpse.

Freshly Killed

Reheated

Difference

Sample size

19

19

19

Mean

417.9

376.1

41.8

Standard Deviation

60

74.7

91.6

SE mean

13.8

17.1

21.0

 There 95% CI for mean difference is (-86,2.3)

T-Value = -1.99

 P-Value = 0.062

Null hypothesis is not rejected, because P- value 0.062 is greater than 0.05 and therefore not significant. Also since the confidence interval contains 0, its not significant. This means there is no difference in cooling rates when rats were killed and reheated.

Parameters

Difference

Sample size (n)

120

Mean

0.130

Standard Error (SED)

0.053

The 95% credible interval for the mean difference is between the interval (0.026, 0.234)

There is 95 % certainty that the difference is between the value of 0.026 and 0.234. Yes, there is the difference.

U.S army recruits in Iraq

Yes, the measurements are still affected by sampling error. When the size of the sample is reduced by random sampling, it increases the sample error.

The recruits are chosen at random hence will reduce biasness, hence the recruit will have better chance of representing the whole population.

The above study is an observational study, because the researcher has no influence on how the fishes fall into groups. Therefore the study is purely observational.

The two variables associated with this study are: subspecies (group) of guppy in Venezuela and most sensitivity wavelength of light.

The explanatory variable in this study is group of guppy in Venezuela and the response variable is wavelength of light.

(In log 10 units)

N

39

39

Mean

5.50

155.50

Standard Deviation 

0.2555

5.36

The test hypothesis associated with chi-square test is:

Ho: Treatments and response are independent

Versus

Ha: Treatments and response are dependent

The chi-square test is given by:

Treatment

Favorable

Unfavorable

Placebo

28.90

35.10

Test drug 

27.10

32.90

Pearson Chi-Square = 21.709, DF = 1, P-Value = 0.000

Likelihood Ratio Chi-Square = 22.377, DF = 1, P-Value = 0.000

Reject the null hypothesis that treatment and response are independent, because the p value is less than 0.05. Hence. treatments and response are dependent.

We need to test the hypothesis

Ho : There is no change of testorone levels vs. Ha : There is a change of testorone values.

After

Before

Difference

Sample size

4

4

4

Mean

88.50

91.50

-3

Standard Deviation

91.50

7.33

2.58

SE mean

2.72

3.66

1.29

95% CI for mean difference: (-7.11, 1.11)

T-Test of mean difference = 0 (vs ≠ 0): T-Value = -2.32  P-Value = 0.103

We don't reject the null hypothesis that "There is no change of testorone levels" because the P. Value =0.103 is more than the significant value(0.05). Also because the confidence interval contains 0 hence its not significant. Hence, There is no change of testorone levels taken before and after watching a game.

Mean

2.26

Median

1.986

Standard Deviation

1.787

Coefficient of variation

79.07%

One tailed test

Two tailed test

t(0.01, 10)

2.764

±3.169

t0.99, 10

-2.764

±0.01285

t0.025, 7

2.365

±2.841

t0.975, 7

-2.365

±0.03248

t0.005, 23

2.807

±3.104

t0.995, 23

-2.807

±0.006335

N

12

Mean

16.69

Standard Deviation

0.704

SE Mean

0.203

The confidence interval for the sample mean is (16.244, 17.139)

There is 95% confidence that the true mean will be between the intervals (16.244, 17.139)

The formulated null hypothesis is:

Ho:  Time taken is the same for both calculators Versus Ha: Time taken is not the same between the calculators.

Sample

Calculator A

Calculator B

Sample size(n)

12

12

Mean

24.67

23.17

Standard Deviation

4.89

4.17

SE Mean

1.4

1.2

Difference = μ (B) - μ (A)

Estimate for difference:  -1.50

95% CI for difference:  (-5.36, 2.36)

T-Test of difference = 0 (vs ≠): T-Value = -0.81  P-Value = 0.428  DF = 21

Don't reject the null hypothesis that time taken is the same for both calculators because P-value is more than significant (0.05). Hence time taken is the same for both calculators to process result.

The test hypothesis associated test is:

Ho: There is no association between hair and color in human beings.

Versus

Ha: There is association between hair and color in human beings.

x

Dark hair

Light hair

Total

Brown eyes

782

241

1023

Blue eyes

234

60

294

Total

1016

301

1317

Dark hair

Light hair

Brown Eyes

789.2

233.8

Blue Eyes 

226.8

67.2

Pearson Chi-Square = 1.285, DF = 1, P-Value = 0.257

Likelihood Ratio Chi-Square = 1.310, DF = 1, P-Value = 0.252

Dont reject the null hypothesis that There is no association between hair and color in human beings because the p-value is more than 0.05. Hence, There is no association between hair color and human beings.