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The highway department is testing two types of reflecting paint for concrete bridge end pillars. The two kinds of paint are alike in every respect except that one other is yellow. The orange paint is applied to 12 bridges, and the yellow paint is applied to 12 bridges. After a period of 1 year, reflectometer readings were made end pillars. (A higher reading means better visibility.) For the orange paint, the mean reflectometer reading was x19.4, with standard deviation s1-2.5. For the mean was X2-6.5, with standard deviation S2-2.4. Based on these data, can we conclude that the yellow paint has less visibility after 1 year?
Use a 10% level What are we testing in this problem?
a. difference of means
b. single proportion
c. difference of proportions
d. single mean
e. paired difference

Sagot :

Answer:

a. difference of means

Step-by-step explanation:

Given that :

Mean , x = 9.4

Standard deviation, [tex]s.d_1[/tex] = 2.5

Number, [tex]n_1[/tex] = 12

Mean, y = 6.5

standard deviation, [tex]s.d_2[/tex] = 2.4

Number, [tex]n_2[/tex] = 12

The null hypothesis is : [tex]$H_0: \mu_1=\mu_2$[/tex]

The alternate hypothesis is : [tex]$H_1: \mu_1>\mu_2$[/tex]

Level of significance, [tex]\alpha[/tex] = 0.1

From the [tex]\text{standard normal table, right tailed,}[/tex] [tex]$t_{1/2}$[/tex] = 1.363

Since out test is right tailed.

Reject [tex]H_0[/tex], if [tex]$T_0>1.363$[/tex]

We use the test statics,

[tex]$t_0=\frac{(x-y)}{\sqrt{\frac{s.d_1}{n_1}+\frac{s.d_2}{n_2}}}$[/tex]

[tex]$t_0=\frac{(9.4-6.5)}{\sqrt{\frac{6.25}{12}+\frac{5.76}{12}}}$[/tex]

[tex]$t_0=2.899$[/tex]

[tex]$|t_0|=2.899$[/tex]

[tex]\text{Critical value}[/tex]

The value of [tex]$|t_{1/2}|$[/tex] with minimum [tex]$\left(n_1-1,n_2-1)$[/tex] that is 11 df is 1.363

We go [tex]$|t_0|=2.899$[/tex] and [tex]$|t_{1/2}|$[/tex] = 1.363

Decision making:

Since the value of [tex]|t_0|>|t_{1/2}|$[/tex]  and we reject the [tex]H_0[/tex]

The p-value : right tail [tex]H_a:(p>2.8988)[/tex]

                                      = 0.00724

Therefore the value of [tex]$p_{0.1} > 0.00724$[/tex], and so we reject the [tex]H_0[/tex]

Thus we are testing 'the difference of means" in this problem.

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