Let [tex]\mu[/tex] be the population mean.
As per given we have,
[tex]H_0:\mu=80\\\\ H_a:\mu>80[/tex] , since alternative hypothesis is right tailed , so the test is a right-tailed test.
Test statistic : [tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For, n=80 [tex]\sigma=12,\ \overline{x}=80[/tex] , we have
[tex]z=\dfrac{83-80}{\dfrac{12}{\sqrt{80}}}\approx2.236[/tex]
Critical z-value at [tex]\alpha=0.05[/tex] =1.645
Since the test statistic value is greater than the critical z-value, so we reject the null hypothesis, i.e. alternative hypothesis is accepted.
Conclusion: We have enough evidence to support the claim that the mean work week has increased for women at the 5% level.