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3 years ago

15

A multiple regression model has ___

1 answer:

Salsk061 [2.6K]3 years ago

5 0

Answer:

C. More than one independent variable

Step-by-step explanation:

The difference between linear regression model and multiple regression model is that the linear regression has just one independent variable which is use to determine the dependent variable. While the multiple regression model has at least two or more independent variable which is use to determine the dependent variable.

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X- 0,1,2,3 Y- 3,4,11,18. Which linear equatio represents the data given in the table? A) y = 3x + 7 B) y = 7x + 3 C) y = 3x - 7

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3. A new process has been developed for applying pho toresist to 125-mm silicon wafers used in manufac turing integrated circu

dlinn [17]

Answer:

a

The null hypothesis is $H_o : \mu = 13.4000$

The alternative hypothesis is $H_a : \mu \ne 13.4000$

The null hypothesis is rejected

b

The 99% confidence level is $13.3930 < \mu < 13.3994$

Step-by-step explanation:

From the question we are told that

The sample size is n = 10

The population mean is $\mu = 13.4 000 \ angstroms$

The level of significance is $\alpha = 0.05$

The sample data is

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

Generally the sample mean is mathematically represented as

$\= x = \frac{13.3987+ 13.3957\cdots +13.4002 }{10}$

=> $\= x = 13.3962$

Generally the sample standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }$

=> $\sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 + (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2 }{10} }$

=> $\sigma =0.0039$

The null hypothesis is $H_o : \mu = 13.4000$

The alternative hypothesis is $H_a : \mu \ne 13.4000$

Generally the test statistics is mathematically represented as

$z = \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }$

=> $z = \frac{13.3962 - 13.4000}{\frac{0.0039}{\sqrt{10} } }$

=> $z = 3.08$

Generally the p-value is mathematically represented as

$p-value = 2P( z > 3.08)$

From the z-table $P(z > 3.08)= 0.001035$

So

$p-value = 2* 0.001035$

$p-value = 0.00207$

So from the obtained value we see that

$p-value < \alpha$

Hence the null hypothesis is rejected

Consider the b question

Given that the confidence level is 99% then the level of significance is

$\alpha = (100 -99)\%$

=> $\alpha = 0.01$

Generally from the normal distribution table critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 2.58 * \frac{0.0039}{\sqrt{10} }$

=> $E = 0.00318$

Generally the 99% confidence interval is mathematically represented as

$13.3962 - 0.00318 < \mu < 13.3962 + 0.00318$

=> $13.3930 < \mu < 13.3994$

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