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3 years ago

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A sample of 65 observations is selected from one population with a population standard deviation of 0.75. The sample mean is 2.6

7. A sample of 50 observations is selected from a second population with a population standard deviation of 0.66. The sample mean is 2.59. Conduct the following test of hypothesis using the 0.08 significance level.
H0 : μ1 ≤ μ2
H1 : μ1 > μ2

a. Is this a one-tailed or a two-tailed test?
b. State the decision rule.
c. Compute the value of the test statistic.
d. What is your decision regarding H0?
e. What is the p-value?

1 answer:

Angelina_Jolie [31]3 years ago

3 0

Answer:

Step-by-step explanation:

The null hypothesis is

H0 : μ1 ≤ μ2

The alternative hypothesis is

H1 : μ1 > μ2

a) it is a one-tailed and also a right tailed test

b) The decision rule is the basis on which the null hypothesis is rejected or not. If 0.08 > p value, reject H0. If 0.08 < p value, do not reject H0.

c) Since the population standard deviations are known, we would use the formula to determine the test statistic(z score)

z = (x1 - x2)/√σ1²/n1 + σ2²/n2

Where

x1 and x2 represents sample means

σ1 and σ2 represents population standard deviations

n1 and n2 represents number of samples

From the information given,

x1 = 2.67

x2 = 2.59

σ1 = 0.75

σ2 = 0.66

n1 = 65

n2 = 50

z = (2.67 - 2.59)/√0.75²/65 + 0.66²/50

= 0.08/√0.00865 + 0.008712

z = 0.61

From the normal distribution table, the probability value corresponding to the z score is 0.73

d) If 0.08 > 0.73, we would reject H0

e) the p-value is 0.73

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The 95% confidence interval would be given by (18.73;23.07)

Step-by-step explanation:

Assuming these data

20 40 22 22 21 21 20 10 20 20

20 13 18 50 20 18 15 8 22 25

22 10 20 22 22 21 15 23 30 12

9 20 40 22 29 19 15 20 20 20

20 15 19 21 14 22 21 35 20 22

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

First we need to find the sample mean with the following formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{50}=20.9$

And in order to find the sample standard deviation we can use the following formula:

$s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar x)^2}{n-1}}=7.65$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=50-1=49$

Since the Confidence is 0.95or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$

Now we have everything in order to replace into formula (1):

$20.9-2.01\frac{7.65}{\sqrt{50}}=18.73$

$20.9+2.01\frac{7.65}{\sqrt{50}}=23.07$

So on this case the 95% confidence interval would be given by (18.73;23.07)

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