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A manufacturer of cordless electric shavers sampled 13 from a​ day's production and found the mean time of continuous usage without recharging to be 410 minutes with a sample standard deviation of 30 minutes. We can assume that times are normally distributed. We wish to test if the true mean operating time without recharging is more than 400 minutes. What are the correct null and alternative​ hypotheses?

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Answer:

The idea is test if the true mean operating time without recharging is more than 400 minutes (alternative hypothesis) and the complement would represent the null hypothesis, the system of hypothesis are then:    

Null hypothesis:[tex]\mu \leq 400[/tex]    

Alternative hypothesis:[tex]\mu > 400[/tex]    

Step-by-step explanation:

Information given

[tex]\bar X=410[/tex] represent the sample mean time of continuous usage without recharging  

[tex]s=30[/tex] represent the population standard deviation  

[tex]n=13[/tex] sample size    

[tex]\mu_o =400[/tex] represent the value to check

[tex]\alpha[/tex] represent the significance level

t would represent the statistic  for the test

[tex]p_v[/tex] represent the p value for the test

Hypothesis to verify

The idea is test if the true mean operating time without recharging is more than 400 minutes (alternative hypothesis) and the complement would represent the null hypothesis, the system of hypothesis are then:    

Null hypothesis:[tex]\mu \leq 400[/tex]    

Alternative hypothesis:[tex]\mu > 400[/tex]    

Since we don't know the population deviation we need to use the t test with the following statistic:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)    

Replacing the info we got:

[tex]z=\frac{410-400}{\frac{30}{\sqrt{13}}}=1.202[/tex]    

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