Question

[12 points] The following data, recorded in days, represent the length of time to recovery for patients randomly treated with one of two medications to c lear up severe bladder infections: Medication2 n, = 14 17 2-16 小1.8 Find a 99 % confidence interval for the difference two medications, assuming normal populations with equal variances in the mean recovery time for the μ2-ui value, conclusion about the null hypothesis, and interpret the result: The measured radiation emissions (in W/kg) for a random sample of 11 cell phones produce a sample mean of 0.938 and a sample standard deviation of 0.423. Use a 0.05 level of significance to test the claim that cell phones have a mean radiation level that is less than 1.00 Wkg. Assume that the sample is from a population that is normally distributed 8. [10 points] In the following identify the null hypothesis, alternative hypothesis, test statistik, critical

Answer 1

SOLUTION:

From given data,

(7) The following data, recorded in days, represents the length of time to recovery for patients randomly treated with one of the two medications to clear up severe bladder infections:

$n_{1}$  = 14	$n_{2}$ = 16
$\bar{x}_{1}$ = 17	$\bar{x}_{2}$ = 19
$s_{1}^{2}$ = 1.5	$s_{2}^{2}$ = 1.8

The 99% confidence interval for the difference  $\mu _{1}$ -
μ2
.

Find the standard error for comparing the means.

Since Variance are assumed equal , but known , we used pooled variance.

The pooled estimate of Variance is:

2
= ($n_{1}$ - 1)$s_{1}^{2}$ + ($n_{2}$ - 1)$s_{2}^{2}$ / ($n_{1}$ - 1) + ($n_{2}$ - 1)

= (14- 1)*1.5² + (16- 1) *1.8² / (14 - 1) + (16- 1)

= 29.25 + 48.6 / 28

= 77.85 / 28

= 2.78

The standard error = SE = sqrt (
2
/$n_{1}$ +
2
/ $n_{2}$ )

= sqrt (2.78 /14 + 2.78 / 16 )

=  0.61018

Standard error = 0.61018

99% confidence interval

99/100 = 0.99

Significance level = $\alpha$ = 1-0.99 = 0.01

$\alpha$/2 = 0.01/2 =0.005

Degree of freedom = df = $n_{1}$ + $n_{2}$ - 2 = 14+16-2 = 28

Critical value

Critical value = t_{$\alpha$/2,df} = t_0.005,28 = 2.763 (from t-table , two -tails, df=28)

Margin of Error

Margin of Error = E = t_{$\alpha$/2,df}$\times$SE

= 2.763  $\times$0.61018

= 2.763 $\times$ 1.68592734

E = 4.658

Point Estimate of difference : $\bar{x_{1}}$ - $\bar{x}_{2}$  = 17 - 19 = -2

Limits of 95% confidence interval are given by :

Lower limit = ($\bar{x_{1}}$ - $\bar{x}_{2}$ )- E =-2-4.658= -6.658

Upper limit =($\bar{x_{1}}$ - $\bar{x}_{2}$ )+ E = -2+4.658= 2.658

99% confidence interval using t-dist:

-6.658 < $\mu _{1}$ -
μ2
< 2.658

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