Answer in Statistics and Probability for Gel #251710

Sinve the population is normally distributed and standart deviation is known, we can use Z-statistic

Null hypothesis: “u = 50”

Alternative hypothesis: “u < 50”

Test statistic: “Z={frac {(u-u{scriptscriptstyle 0})*sqrt{n}} {u03c3}}”, where u – sample mean, “u{scriptscriptstyle 0}” – claimed mean, n – sample size, σ – standard deviation

In our case: “Z={frac {(42-50)*sqrt{35}} {11.9}} = -3.98”

Due to the form of the alternative hypothesis, left-tailed test is appropriate

The critical value can be found as “P(N(0,1)<Z{scriptscriptstyle {cr}}) = alpha”, where “alpha” – level of significance, “Z{scriptscriptstyle {cr}}” – critical value

In our case: “P(N(0,1)<Z{scriptscriptstyle {cr}}) = 0.01 to Z{scriptscriptstyle {cr}}” = – 2.33

Since “Z” < “Z{scriptscriptstyle {cr}}”, we reject the null hypothesis.

There are statistically significant evidence that the population mean is now less than 50