A report in 2019 showed that 65.2% of 5500 internet users use google chrome as their primary web browser. Then, the report in 2020 showed that 67% of 2800 internet users use google chrome as their primary web browser. Can you conclude use significance level of 0.01 that the proportion of internet users use google chrome as their primary web browser in 2019 is less than the proportion in 2020? If we let:Let μ1 be the population mean number of internet users use google chrome as their primary web browser in 2019, and μ2 be the population mean number of internet users use google chrome as their primary web browser in 2020.Let p1 be the population proportion of internet users use google chrome as their primary web browser in 2019, and p2 be the population proportion of internet users use google chrome as their primary web browser in 2020.Round to four decimal places if necessary. a) Use "mu1" for μ1, "mu2" for μ2, "p1" for p1 and "p2" for p2. Null hypothesis H0 : Alternate hypothesis H1 : b) Type of test: left or right or twoc) significance level: α = d) Test statistic: Clearly state whether test statistic in this claim is z or t. For example, "z=1.2345"e) Compute p-value of the test statistic. f) Decision: Type "yes" if reject null hypothesis. Type "no" if not to reject null hypothesis.

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MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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A report in 2019 showed that 65.2% of 5500 internet users use google chrome as their primary web browser. Then, the report in 2020 showed that 67% of 2800 internet users use google chrome as their primary web browser. Can you conclude use significance level of 0.01 that the proportion of internet users use google chrome as their primary web browser in 2019 is less than the proportion in 2020?

If we let:

Let μ₁ be the population mean number of internet users use google chrome as their primary web browser in 2019, and μ₂ be the population mean number of internet users use google chrome as their primary web browser in 2020.

Let p₁ be the population proportion of internet users use google chrome as their primary web browser in 2019, and p₂ be the population proportion of internet users use google chrome as their primary web browser in 2020.

Round to four decimal places if necessary.

a) Use "mu1" for μ₁, "mu2" for μ₂, "p1" for p₁ and "p2" for p₂.

Null hypothesis H₀ :

Alternate hypothesis H₁ :

b) Type of test: left or right or two

c) significance level: α =

d) Test statistic: Clearly state whether test statistic in this claim is z or t. For example, "z=1.2345"

e) Compute p-value of the test statistic.

f) Decision: Type "yes" if reject null hypothesis. Type "no" if not to reject null hypothesis.

Expert Solution

Step 1

Given Information:

2019 Sample:

Sample size ( $n_{1}$ ) = 5500

Sample proportion ( ${\hat{p}}_{1}$ ) = 65.2% = 0.652

2020 sample:

Sample size ( $n_{2}$ ) = 2800

Sample proportion ( ${\hat{p}}_{2}$ ) = 67% = 0.67

Significance level $α = 0.01$

(a) State the Hypothesis as follows:

Null Hypothesis $H_{0} : p_{1} \geq p_{2}$ (or) $H_{0} : p_{1} = p_{2}$ i.e., population proportion of internet users use google chrome as their primary web browser in 2019 is greater than or equal to the proportion in 2020.

Alternative Hypothesis $H_{1} : p_{1} < p_{2}$ i.e., proportion of internet users use google chrome as their primary web browser in 2019 is less than the proportion in 2020

Step 2

(b) Since, the alternative hypothesis includes "<" operator. This is a Left tailed test.

(d) Test statistic is given by the formula:

$z = \frac{{\hat{p}}_{1} - {\hat{p}}_{2}}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_{1}} + \frac{1}{n_{2}})}}$

where,

$\begin{array}{rcl} \hat{p} & = & \frac{n_{1} {\hat{p}}_{1} + n_{2} {\hat{p}}_{2}}{n_{1} + n_{2}} \\ = & \frac{5500 \times 0.652 + 2800 \times 0.67}{5500 + 2800} \\ = & \frac{3586 + 1876}{8300} \\ = & 0.6580722892 \\ \approx & 0.6581 \end{array}$

Substitute the values in the formula of the test statistic:

$\begin{array}{rcl} z & = & \frac{{\hat{p}}_{1} - {\hat{p}}_{2}}{\sqrt{\hat{p} (1 - \hat{p}) (\frac{1}{n_{1}} + \frac{1}{n_{2}})}} \\ = & \frac{0.652 - 0.67}{\sqrt{0.6581 (1 - 0.6581) (\frac{1}{5500} + \frac{1}{2800})}} \\ = & \frac{- 0.018}{\sqrt{0.22500439 (0.0001818 + 0.0003571)}} \\ = & \frac{- 0.018}{0.0110116} \\ \approx & - 1.6346 \end{array}$

Thus, the value of the test statistic is -1.6346

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