Answer to Question #187024 in Statistics and Probability for Cherry Blythe

Question

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Perform Hypothesis Testing

1. A factory manufacturing light-emitting diode (LED) bulbs claims that their light bulbs last for 50,000 hours on the average. To confirm if this claim is valid, a quality control manager got a sample of 50 LED bulbs and obtained a mean lifespan of 40,000 hours. The standard deviation of the manufacturing process is 1000 hours. Do you think that the claim of the manufacturer is valid at the 5% level of significance?
2. A presidential candidate asks a polling organization to conduct a nationwide survey to determine the percentage of potential voters who would vote for him over his rival presidential candidate. Out of 2500 respondents in the sample, 925 said they would vote for him. If 40% of the potential voters vote for his rival, is this significantly different from the percentage of potential voters of the candidate who requested the survey? Use 5% level of significance.

Answer 1

1.) We have given that,

$\bar x = 40000$

$s= 1000$

$n = 50$

Degree of freedom = 50-1 = 49

$\alpha = 0.05$

The following null and alternative hypothesis needed to be tested:

$H_0 : \mu = 50000$

$H_1 : \mu \ne 50000$

We will use t test in this scenario.

Based on the information provided, the critical value for a two tailed test is $t_c = 2.009$

The t statistic is calculated as

$t = \dfrac{\bar x - \mu}{\dfrac{s}{\sqrt n}}$

$= \dfrac{40000-50000}{\dfrac{1000}{\sqrt 50}}$

$= -70.71$

Hence, $70.711 > 2.009 = t_c$ it is concluded that null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean is different than 50000, at 0.05 level of significance.

2.) Proportion of people who vote for him $= p = \dfrac{925}{2500} = 0.37$

Hence, we can say that this significantly different from the percentage of potential voters of the candidate who requested the survey