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Examine the data to see if there is enough evidence to establish that the four brands have different average battery life.

The null and alternative hypotheses should be stated.

Hypothesis of nullity:

$H_0:{渭}_1={渭}_2={渭}_3={渭}_4$

That is, no variation in average battery life exists across the four brands.

$H_a$alternative hypothesis: At least one ${渭}_i$is different from the rest.

That is to say, the four brands have different average battery life.

Determine the significance level.

The importance level here is, $伪=0.05$

Calculate the test statistic's value.

Procedure in Excel:

Step 1: Select $\mathrm{Data}>\mathrm{Data}\mathrm{Analysis}>\mathrm{Anova}:\mathrm{Single}\mathrm{Factor}$from the menu bar.

Step 2: Select $\mathrm{A}1:\mathrm{D}11$from the Input range.

Step 3: Press the $\mathrm{OK}$button.

Anova: Single Factor(Excel Output)

The value of $\mathrm{F}$ in the Excel output is $6.602$.

The crucial number is $2.9223$, according to the Excel result.

The data-custom-editor="chemistry" $\mathrm{P}$-value is $0.00147$, according to the Excel output.

The value of the test statistic is now in the rejection zone. Specifically, $F(=6.602)>F_{crit}(=2.9223)$. As a result, the null hypothesis is rejected at the$5\%$level.

As a result, the outcome is statistically significant.

P-value method:

Rule of rejection:

If $P鈮�伪$, the null hypothesis must be rejected.

The $\mathrm{P}$-value is $0.00147$, which is less than the significance level. Specifically, $P(=0.00147)<alpha(=0.05)$. As a result, the null hypothesis is rejected at the$5\%$level.

As a result, the test results are statistically significant at a level of significance of $5$percent.

Interpretation:

The statistics are sufficient to infer that at the $5percent$level, there is a difference in mean battery life between the four brands.