A company has developed a new deluxe AAA battery that is supposed to last longer than its regular AAA battery. However these new batteries are more expensive to produce. So the company would like to be convinced that they really do last longer. Based on years of experience, the company knows that its regular AAA batteries last lor 45 hours of continuous use. On average. The company selects an SRS Of 50 new batteries and uses them continuously until they are completely drained. The Sample mean lifetime is X =46.9 hours with a Standard deviation of S=4.6 hours A) Check for the conditions for the situation B) Calculate the test statistic for this situation C) What is the P value for this situation? D) What conclusion do you draw with 5% significance level? why? E) What type of error could you possibly make here? I

[tex]\mu=the\text{ true mean lifetime of new AAA batteries}[/tex][tex]\begin{gathered} H_0\colon\mu=45\text{ hours} \\ H_a\colon\mu>45\text{ hours} \end{gathered}[/tex][tex]n=50[/tex]B)

Calculating t test statistic

[tex]\begin{gathered} t=\frac{statistic-parameter}{s\tan dard\text{ deviation of statistic}} \\ t=\frac{\bar{x}-\mu_0}{\frac{s_x}{\sqrt[]{n}}} \end{gathered}[/tex]

Plugging in the values, we have:

[tex]\begin{gathered} t=\frac{\bar{x}-\mu_0}{\frac{s_x}{\sqrt[]{n}}} \\ t=\frac{46.9-45}{\frac{4.6}{\sqrt[]{50}}} \\ t=\frac{1.9}{0.6505} \\ t=2.9207 \end{gathered}[/tex]C)

t test statistic = 2.9207

degrees of freedom = n - 1 = 50 - 1 = 49

Using a calculator, we can calculate the p-value.

[tex]p-\text{value}=0.002633[/tex]D)

Since p value is less than significance level (p value < alpha), then we will reject H_0 and take the alternate hypothesis.

Thus the test suggests that the new batteries do last more than 45 hours.

E)

We could've done Type I error here.

Type I error or α: Reject the null when it’s true.