Question 1200934
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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<ul><li>We were given the value of sigma.</li><li>The sample size is larger than 30. When n > 30, the T distribution starts to look like the Z distribution (approximately). Refer to the <a href = "https://www.statology.org/central-limit-theorem/">Central Limit Theorem</a> for more information. This <a href = "https://www.statology.org/large-sample-condition/">relevant article</a> is also something I recommend for further reading.</li></ul>Here's a flowchart to help remember when to use either test.
<img width=500 src = "https://miro.medium.com/v2/resize:fit:1100/format:webp/1*XLhQMDdKW-3sd8a8HSFvgA.png">
Image Source
<a href = "https://towardsdatascience.com/introduction-tfrom-the-central-limit-theorem-to-the-z-and-t-distributions-66513defb175">https://towardsdatascience.com/introduction-tfrom-the-central-limit-theorem-to-the-z-and-t-distributions-66513defb175</a>


Now use a table such as this one
<a href = "https://www.ztable.net/">https://www.ztable.net/</a>
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 <font color=red>p-value (0.07672) is NOT smaller than alpha</font> = 0.05, we <font color=red>fail to reject the null</font>.


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 <font color=blue>The sample battery life is not significantly different from 900 days</font>
In other words,<font color=blue>The average lifespan of a cellphone battery appears to be 900 days</font>


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