SOLUTION: A mobile phone company is concerned about the lifetime of phone batteries supplied by a new supplier. Based upon historical data this type of battery should last for 900 days with

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Question 1200934: A mobile phone company is concerned about the lifetime of phone batteries supplied by a new supplier. Based upon historical data this type of battery should last for 900 days with a standard deviation of 150 days. A recent randomly selected sample of 40 batteries was selected and the sample battery life was found to be 942 days. Is the sample battery life significantly different from 900 days (significance level 5%)?
Ans: (Ho is accepted.) Am I right?
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

mu = population mean
sigma = population standard deviation
n = sample size
xbar = sample mean

Hypothesis:
Ho: mu = 900
Ha: mu =/= 900
The "not equals" sign in the alternative hypothesis tells us that we're doing a two-tailed test.

Other given info
sigma = 150
n = 40
xbar = 942

Compute the test statistic.
z = (xbar - mu)/(sigma/sqrt(n))
z = (942 - 900)/(150/sqrt(40))
z = 1.7708754896943
z = 1.77
Notice I used the Z statistic instead of T statistic.
This is valid for two reasons
  • We were given the value of sigma.
  • The sample size is larger than 30. When n > 30, the T distribution starts to look like the Z distribution (approximately). Refer to the Central Limit Theorem for more information. This relevant article is also something I recommend for further reading.
Here's a flowchart to help remember when to use either test.

Image Source
https://towardsdatascience.com/introduction-tfrom-the-central-limit-theorem-to-the-z-and-t-distributions-66513defb175

Now use a table such as this one
https://www.ztable.net/
to find that
P(Z < 1.77) = 0.96164
Or you can use a stats calculator for better accuracy.

This then leads to
P(Z > 1.77) = 1 - P(Z < 1.77)
P(Z > 1.77) = 1 - 0.96164
P(Z > 1.77) = 0.03836
This is the approximate area under the standard normal curve to the right of z = 1.77

Because we're doing a two-tailed test, we'll double that result
2*0.03836 = 0.07672
This represents the p-value.

Rule: If the p-value is smaller than alpha, then we reject the null.
A handy memorization phrase could be "If the p-value is low, then the null must go".

Since the p-value (0.07672) is NOT smaller than alpha = 0.05, we fail to reject the null.

Therefore, the null is "accepted". I put that in quotes because technically we haven't accepted it. Rather we just haven't found enough evidence to reject it, so we have no choice but to side with the null.

Your answer of "Ho is accepted" is correct.
One interpretation would be The sample battery life is not significantly different from 900 days
In other words,The average lifespan of a cellphone battery appears to be 900 days

Side notes:
  • 900 days = 900/365 = 2.4658 years (approximate)
  • 0.4658 years = 0.4658*365 = 170.017 = 170 days (approximate)
  • 2.4658 years = 2 years + 0.4658 years = 2 years + 170 days (approximate)
  • Therefore, 900 days = 2 years + 170 days (approximate)
  • This is ignoring leap years.