"\\bar x=990"
"n=100"
"\\mu = 1000"
This is a test of a single population mean and the population standard deviation is known and the sample size is greater than 30 so this is a z-test.
Then looking up the calculated Z-score in a table of the standard normal distribution cumulative probability we find the corresponding probability or p-value.
a) "\\sigma=25"
"Z_{test}=\\frac{990-1000}{\\frac{25}{\\sqrt {100}}}=-4 \\implies" p-value = 0.000032
b) "\\sigma=50"
"Z_{test}=\\frac{990-1000}{\\frac{50}{\\sqrt {100}}}=-2 \\implies" p-value = 0.02275
c) "\\sigma=100"
"Z_{test}=\\frac{990-1000}{\\frac{100}{\\sqrt {100}}}=-1 \\implies" p-value = 0.158655
d) If the standard deviation increases the value of the test statistic value decreases and the p-value increases.
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