Let [tex]\mu[/tex] be the population mean.
Null hypothesis : [tex]H_0:\mu\geq26[/tex]
Alternative hypothesis : [tex]H_1:\mu<26[/tex]
Since the alternative hypothesis is left tailed, so the test is a left-tailed test.
Sample size : n=5 <30 , so we use t-test.
Test statistic: [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]t=\dfrac{25.2-26}{\dfrac{1}{\sqrt{5}}}\approx-1.79[/tex]
Critical t-value for t=[tex]t_{n-1, \alpha}=t_{4,0.06}=1.9712[/tex]
Since, the absolute value of t (1.79) is less than the critical t-value , so we fail to reject the null hypothesis.
Hence, we have sufficient evidence to support the company's claim.